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Soal Notasi Sigma dan Pembahasan

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Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) berbagi Kumpulan Soal Notasi Sigma dan Pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.
Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cek jawaban kamu dengan pembahasan yang telah disediakan, dengan cara:
klik "LIHAT/TUTUP:".
Soal No. 1
Nilai dari $\sum\limits_{n=3}^{6}{(12-n^2)}$ = ….
A. $-41$
B. $-38$
C. $-35$
D. $-30$
E. $-24$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=3}^{6}{(12-n^2)}$
= $(12-3^2)+(12-4^2)+(12-5^2)+(12-6^2)$
= $3+(-4)+(-13)+(-24)$
= $-38$
Jawaban: B
Soal No. 2
Nilai dari $\sum\limits_{p=2}^{5}{\frac{p}{p+1}}$ = ….
A. 3,05
B. 3,10
C. 3,20
D. 3,25
E. 3,50
Penyelesaian: Lihat/Tutup $\sum\limits_{p=2}^{5}{\frac{p}{p+1}}$
= $\frac{2}{2+1}+\frac{3}{3+1}+\frac{4}{4+1}+\frac{5}{5+1}$
= $\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}$
= $\frac{2\times 20}{3\times 20}+\frac{3\times 15}{4\times 15}+\frac{4\times 12}{5\times 12}+\frac{5\times 10}{6\times 10}$
= $\frac{40+45+48+50}{60}$
= 3,05
Jawaban: A
Soal No. 3
Notasi sigma yang menyatakan penjumlahan 20 bilangan asli ganjil adalah ….
A. $\sum\limits_{n=1}^{20}{1}$
B. $\sum\limits_{n=1}^{20}{(n-1}$
C. $\sum\limits_{n=1}^{20}{(2n-1)}$
D. $\sum\limits_{n=1}^{10}{(n-1)}$
E. $\sum\limits_{n=1}^{10}{(2n-1)}$
Penyelesaian: Lihat/Tutup $\underbrace{1+3+5+7+...}_{20\,bilangan\,ganjil\,pertama}$
= $(2.1-1)+(2.2-1)+(2.3-1)+...+(2.20-1)$
= $\sum\limits_{n=1}^{20}{(2n-1)}$
Jawaban: C
Soal No. 4
Notasi sigma yang menyatakan penjumlahan 25 bilangan kuadrat pertama adalah ….
A. $\sum\limits_{n=1}^{5}{2n}$
B. $\sum\limits_{n=1}^{5}{n^2}$
C. $\sum\limits_{n=1}^{25}{2n}$
D. $\sum\limits_{n=1}^{25}{n^2}$
E. $\sum\limits_{n=1}^{625}{n^2}$
Penyelesaian: Lihat/Tutup $\underbrace{1^2+2^2+3^2+...}_{25\,bilangan\,kuadrat\,pertama}$
= $1^2+2^2+3^2+...+25^2$
= $\sum\limits_{n=1}^{25}{n^2}$
Jawaban: D
Soal No. 5
Diketahui $\sum\limits_{i=1}^{10}{a_i}=40$ dan $\sum\limits_{i=1}^{10}{b_i}=50$, maka $\sum\limits_{i=1}^{10}{(4a_i-b_i+2)}$ = ….
A. 100
B. 110
C. 120
D. 130
E. 140
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{10}{(4a_i-b_i+2)} &= 4\sum\limits_{i=1}^{10}{a_i}-\sum\limits_{i=1}^{10}{b_i}+\sum\limits_{i=1}^{10}{2} \\ &= 4.40-50+10.2 \\ &= 160-50+20 \\ &= 130 \end{align}$
Jawaban: D
Soal No. 6
Diketahui $\sum\limits_{k=5}^{25}{(2-pk)}=0$, maka nilai $\sum\limits_{k=5}^{25}{pk}$ = ….
A. 20
B. 28
C. 30
D. 42
E. 112
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=5}^{25}{(2-pk)} &= 0 \\ \sum\limits_{k=5}^{25}{2}-\sum\limits_{k=5}^{25}{pk} &= 0 \\ \sum\limits_{k=5}^{25}{2} &= \sum\limits_{k=5}^{25}{pk} \\ \sum\limits_{k=5-4}^{25-4}{2} &= \sum\limits_{k=5}^{25}{pk} \\ \sum\limits_{1}^{21}{2} &= \sum\limits_{k=5}^{25}{pk} \\ 21\times 2 &= \sum\limits_{k=5}^{25}{pk} \\ 42 &= \sum\limits_{k=5}^{25}{pk} \end{align}$
Jawaban: D
Soal No. 7
Jika bentuk $\sum\limits_{x=4}^{20}{(3-4x)}$ diubah dengan notasi sigma lain yang sesuai dengan batas bawah satu, akan diperoleh bentuk ….
A. $\sum\limits_{x=1}^{20}{(9-4x)}$
B. $\sum\limits_{x=1}^{20}{(3-4x)}$
C. $\sum\limits_{x=1}^{20}{(-9-4x)}$
D. $\sum\limits_{x=1}^{17}{(9-4x)}$
E. $\sum\limits_{x=1}^{17}{(-9-4x)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{x=4}^{20}{(3-4x)} &= \sum\limits_{x=4-3}^{20-3}{(3-4(x+3))} \\ &= \sum\limits_{x=1}^{17}{(3-4x-12)} \\ &= \sum\limits_{x=1}^{17}{(-9-4x)} \end{align}$
Jawaban: E
Soal No. 8
Jika batas bawah diubah menjadi tujuh, bentuk notasi sigma $\sum\limits_{n=3}^{7}{(6-3n)}$ akan menjadi ….
A. $\sum\limits_{n=7}^{11}{(6-3n)}$
B. $\sum\limits_{n=7}^{11}{(9-3n)}$
C. $\sum\limits_{n=7}^{11}{(18-3n)}$
D. $\sum\limits_{n=7}^{11}{(24-3n)}$
E. $\sum\limits_{n=7}^{11}{(36-3n)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=3}^{7}{(6-3n)} &= \sum\limits_{n=3+4}^{7+4}{(6-3(n-4))} \\ &= \sum\limits_{n=7}^{11}{(6-3n+12)} \\ &= \sum\limits_{n=7}^{11}{(18-3n)} \end{align}$
Jawaban: C
Soal No. 9
$\sum\limits_{i=1}^{n}{({{i}^{2}}+5i)}$ = ….
A. $\frac{n(n+1)(n+3)}{6}$
B. $\frac{n(n+1)(n+6)}{6}$
C. $\frac{n(n+1)(2n+3)}{6}$
D. $\frac{n(n+1)(2n+12)}{6}$
E. $\frac{n(n+1)(2n+16)}{6}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{n}{({{i}^{2}}+5i)}$
= $\sum\limits_{i=1}^{n}{{{i}^{2}}}+5\sum\limits_{i=1}^{n}{i}$
= $\frac{n(n+1)(2n+1)}{6}+5.\frac{n(n+1)}{2}$
= $\frac{n(n+1)(2n+1)}{6}+\frac{15n(n+1)}{6}$
= $\frac{n(n+1)(2n+1)+15n(n+1)}{6}$
= $\frac{n(n+1)[2n+1+15]}{6}$
= $\frac{n(n+1)(2n+16)}{6}$
Jawaban: E
Soal No. 10
Jika $\sum\limits_{i=1}^{n}{(2i+3)}=60$ maka $n$ = ….
A. 4
B. 5
C. 6
D. 7
E. 8
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{n}{(2i+3)} &= 60 \\ 2\sum\limits_{i=1}^{n}{i}+\sum\limits_{i=1}^{n}{3} &= 60 \\ 2.\frac{n(n+1)}{2}+n.3 &= 60 \\ n(n+1)+3n &= 60 \\ n^2+n+3n-60 &= 0 \\ n^2+4n-60 &= 0 \\ (n+10)(n-6) &= 0 \end{align}$
$n+10=0\to n=-10$ (tidak memenuhi)
$n-6=0\to n=6$ (memenuhi)
Jawaban: C
Soal No. 11
$\sum\limits_{k=1}^{40}{3k}+\sum\limits_{k=1}^{40}{(k+5)}$ = ….
A. 3460
B. 3465
C. 3470
D. 3475
E. 3480
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{40}{3k}+\sum\limits_{k=1}^{40}{(k+5)} &= \sum\limits_{k=1}^{40}{(3k+(k+5))} \\ &= \sum\limits_{k=1}^{40}{(4k+5)} \\ &= 4\sum\limits_{k=1}^{40}{k}+\sum\limits_{k=1}^{40}{5} \\ &= 4.\frac{40(40+1)}{2}+40.5 \\ &= 3480 \end{align}$
Jawaban: E
Soal No. 12
Nilai dari $\sum\limits_{k=1}^{110}{2k}+\sum\limits_{k=1}^{110}{(k+1)}$ adalah ….
A. 37290
B. 36850
C. 18645
D. 18425
E. 18420
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{110}{2k}+\sum\limits_{k=1}^{110}{(k+1)} &= \sum\limits_{k=1}^{110}{(2k+k+1)} \\ &= \sum\limits_{k=1}^{110}{(3k+1)} \\ &= 3\sum\limits_{k=1}^{110}{k}+\sum\limits_{k=1}^{110}{1} \\ &= 3.\frac{110(110+1)}{2}+110 \\ &= 18425 \end{align}$
Jawaban: D
Soal No. 13
Jika $p=\sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}$, maka nilai dari $4p$ = ….
A. 1
B. 2
C. 3
D. 4
E. 6
Penyelesaian: Lihat/Tutup $p=\sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}\,....\,(1)$
Dari persamaan (1):
$\begin{align}p &= 3\sum\limits_{n=1}^{\infty }{\frac{n}{3^{n+1}}} \\ \frac{1}{3}p &= \sum\limits_{n=1}^{\infty }{\frac{n}{3^{n+1}}} \\ &= \sum\limits_{n=1+1}^{\infty +1}{\frac{n}{{{3}^{(n-1)+1}}}} \\ \frac{p}{3} &= \sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}p-\frac{1}{3}p &= \sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}-\sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}} \\ \frac{2}{3}p &= \frac{1}{3}+\sum\limits_{n=2}^{\infty }{\frac{n}{3^n}}-\sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}} \\ &= \frac{1}{3}+\sum\limits_{n=2}^{\infty }{\frac{1}{3^n}} \\ &= \frac{1}{3}+\left( \frac{1}{9}+\frac{1}{27}+\frac{1}{81}+... \right) \\ &= \frac{1}{3}+\frac{1/9}{1-\frac{1}{3}} \\ &= \frac{1}{3}+\frac{1}{9}.\frac{3}{2} \\ &= \frac{1}{3}+\frac{1}{6} \\ \frac{2}{3}p &= \frac{1}{2} \\ p &= \frac{1}{2}\times \frac{3}{2} \\ p &= \frac{3}{4} \end{align}$
$4p=4.\frac{3}{4}=3$
Jawaban: C
Soal No. 14
$\sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}$ = ….
A. $\frac{1}{3}$
B. $\frac{4}{9}$
C. $\frac{5}{9}$
D. $\frac{9}{4}$
E. $\frac{9}{5}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}$
$\begin{align}S &= 4\sum\limits_{i=1}^{\infty }{\frac{i}{4^{i+1}}} \\ \frac{1}{4}S &= \sum\limits_{i=1}^{\infty }{\frac{i}{4^{i+1}}} \\ &= \sum\limits_{i=1+1}^{\infty }{\frac{i-1}{{{4}^{(i-1)+1}}}} \\ \frac{S}{4} &= \sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}}\,....\,(2) \end{align}$
Kurangkan persamaan (1) dan (2) maka:
$\begin{align}S-\frac{1}{4}S &= \sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}-\sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}} \\ &= \frac{1}{4}+\sum\limits_{i=2}^{\infty }{\frac{i}{4^i}}-\sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}} \\ &= \frac{1}{4}+\sum\limits_{i=2}^{\infty }{\frac{1}{4^i}} \\ &= \frac{1}{4}+\left( \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+... \right) \\ &= \frac{1}{4}+\frac{\frac{1}{16}}{1-\frac{1}{4}} \\ &= \frac{1}{4}+\frac{1}{16}.\frac{4}{3} \\ &= \frac{1}{4}+\frac{1}{12} \\ \frac{3}{4}S &= \frac{1}{3} \\ S &= \frac{4}{9} \end{align}$
Jawaban: B
Soal No. 15
$\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}}$ = ….
A. 8
B. 7
C. 6
D. 5
E. 4
Penyelesaian: Lihat/Tutup $\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}}$
= $\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}\times \frac{\sqrt{2k}-\sqrt{2k-2}}{\sqrt{2k}-\sqrt{2k-2}}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k}-\sqrt{2k-2}}{2k-(2k-2)}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k}-\sqrt{2k-2}}{2}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k-2}-\sqrt{2k}}{-2}}$
= $\frac{\sqrt{4}-\sqrt{6}}{-2}+\frac{\sqrt{6}-\sqrt{8}}{-2}+\frac{\sqrt{8}-\sqrt{10}}{-2}+...+\frac{\sqrt{96}-\sqrt{98}}{-2}+\frac{\sqrt{98}-\sqrt{100}}{-2}$
= $\frac{\sqrt{4}-\sqrt{6}+\sqrt{6}-\sqrt{8}+\sqrt{8}-\sqrt{10}+...+\sqrt{96}-\sqrt{98}+\sqrt{98}-\sqrt{100}}{-2}$
= $\frac{\sqrt{4}-\sqrt{100}}{-2}$
= $\frac{2-10}{-2}$
= 4
Jawaban: E
Soal No. 16
Akar-akar persamaan kuadrat $x^2\sum\limits_{i=1}^{2}{2}+x.\sum\limits_{i=1}^{5}{i}-\frac{2}{3}\sum\limits_{i=1}^{3}{i}=0$ adalah $x_1$ dan $x_2$ dengan $x_1 < x_2$. Nilai dari $x_1+8x_2$ = ….
A. $-2$
B. $-1$
C. 0
D. 2
E. 4
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{2}{2}=2.2=4$
$\sum\limits_{i=1}^{5}{i}=\frac{5(5+1)}{2}=15$
$\sum\limits_{i=1}^{3}{i}=\frac{3(3+1)}{2}=6$
$\begin{align}x^2\sum\limits_{i=1}^{2}{2}+x.\sum\limits_{i=1}^{5}{i}-\frac{2}{3}\sum\limits_{i=1}^{3}{i} &= 0 \\ x^2.4+x.15-\frac{2}{3}.6 &= 0 \\ 4x^2+15x-4 &= 0 \\ (4x-1)(x+4) &= 0 \end{align}$
$4x-1=0\to x=\frac{1}{4}$
$x+4=0\to x=-4$
$x_1 < x_2$ maka $x_1=-4$ dan $x_2=\frac{1}{4}$
$x_1+8x_2=-4+8.\frac{1}{4}=-2$
Jawaban: A
Soal No. 17
Hasil dari $\sum\limits_{p=1}^{5}{\left( \sin \frac{p\pi }{2}+\cos \frac{p\pi }{2} \right)}$ = …
A. $-1$
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1
Penyelesaian: Lihat/Tutup $\sum\limits_{p=1}^{5}{\left( \sin \frac{p\pi }{2}+\cos \frac{p\pi }{2} \right)}$
= $\left( \sin \frac{\pi }{2}+\cos \frac{\pi }{2} \right)$ + $\left( \sin \pi +\cos \pi \right)$ + $\left( \sin \frac{3\pi }{2}+\cos \frac{3\pi }{2} \right)$ + $\left( \sin 2\pi +\cos 2\pi \right)$ + $\left( \sin \frac{5\pi }{2}+\cos \frac{5\pi }{2} \right)$
= $\left( 1+0 \right)$ + $\left( 0-1 \right)$ + $\left( -1+0 \right)$ + $\left( 0+1 \right)$ + $\left( 1+0 \right)$
= 1
Jawaban: E
Soal No. 18
Rumus $\sum\limits_{j=1}^{n}{(j+2)(j-5)}$ = ….
A. $\frac{n(n^2-3n-34)}{2}$
B. $\frac{n(n^2-3n-34)}{3}$
C. $\frac{n(n^2-3n-34)}{4}$
D. $\frac{n(n^2-3n-34)}{5}$
E. $\frac{n(n^2-3n-34)}{6}$
Penyelesaian: Lihat/Tutup $\sum\limits_{j=1}^{n}{(j+2)(j-5)}$
= $\sum\limits_{j=1}^{n}{(j^2-5j+2j-10)}$
= $\sum\limits_{j=1}^{n}{(j^2-3j-10)}$
= $\sum\limits_{j=1}^{n}{j^2}-3\sum\limits_{j=1}^{n}{j}-\sum\limits_{j=1}^{n}{10}$
= $\frac{1}{6}n(n+1)(2n+1)$$-3.\frac{1}{2}n(n+1)$$-10n$
= $\frac{(n^2+n)(2n+1)}{6}$$-\frac{3(n^2+n)}{2}$$-10n$
= $\frac{2n^3+3n^2+n}{6}$$-\frac{3n^2+3n}{2}$$-10n$
= $\frac{2n^3+3n^2+n}{6}-\frac{9n^2+9n}{6}-\frac{60n}{6}$
= $\frac{2n^3-6n^2-68n}{6}$
= $\frac{n^3-3n^2-34n}{3}$
= $\frac{n(n^2-3n-34)}{3}$
Jawaban: B
Soal No. 19
Nilai dari $\sum\limits_{n=5}^{10}{(2n-1)}$ = ….
A. 110
B. 94
C. 84
D. 76
E. 74
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=5}^{10}{(2n-1)} &= \sum\limits_{n=5-4}^{10-4}{(2(n+4)-1)} \\ &= \sum\limits_{n=1}^{6}{(2n+7)} \\ &= 2\sum\limits_{n=1}^{6}{n}+\sum\limits_{n=1}^{6}{7} \\ &= 2.\frac{6(6+1)}{2}+6.7 \\ &= 84 \end{align}$
Jawaban: C
Soal No. 20
$\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}$ = ….
A. $\frac{7}{8}$
B. $\frac{6}{8}$
C. $\frac{5}{8}$
D. $\frac{4}{8}$
E. $\frac{3}{8}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}\,....\,(1)$
$\begin{align}S &= 5\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^{n+1}}} \\ \frac{1}{5}S &= \sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^{n+1}}} \\ &= \sum\limits_{n=1+1}^{\infty +1}{\frac{2(n-1)+1}{{{5}^{(n-1)+1}}}} \\ \frac{1}{5}S &= \sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}S-\frac{1}{5}S &= \sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}-\sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}} \\ \frac{4}{5}S &= \frac{3}{5}+\sum\limits_{n=2}^{\infty }{\frac{2n+1}{5^n}}-\sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}} \\ &= \frac{3}{5}+\sum\limits_{n=2}^{\infty }{\frac{2}{5^n}} \\ &= \frac{3}{5}+2\sum\limits_{n=2}^{\infty }{\frac{1}{5^n}} \\ &= \frac{3}{5}+2\left( \frac{1}{25}+\frac{1}{125}+\frac{1}{625}+... \right) \\ &= \frac{3}{5}+2.\frac{\frac{1}{25}}{1-\frac{1}{5}} \\ &= \frac{3}{5}+2.\frac{1}{25}.\frac{5}{4} \\ &= \frac{3}{5}+\frac{1}{10} \\ \frac{4}{5}S &= \frac{7}{10} \\ S &= \frac{7}{8} \end{align}$
Jawaban: A
Soal No. 21
Notasi sigma yang menyatakan penjumlahan seribu bilangan asli pertama adalah ….
A. $\sum\limits_{n=1}^{1.000}{n}$
B. $\sum\limits_{n=1}^{1.000}{i}$
C. $\sum\limits_{n=1}^{1.000}{1}$
D. $\sum\limits_{n=1}^{999}{n}$
E. $\sum\limits_{n=1}^{999}{1}$
Penyelesaian: Lihat/Tutup $1+2+3+...+1000=\sum\limits_{n=1}^{1000}{n}$
Jawaban: A
Soal No. 22
Notasi sigma yang menyetakan penjumlahan 16 bilangan asli genap pertama adalah ….
A. $\sum\limits_{n=1}^{8}{2}$
B. $\sum\limits_{n=1}^{8}{2n}$
C. $\sum\limits_{n=1}^{16}{2}$
D. $\sum\limits_{n=1}^{16}{n}$
E. $\sum\limits_{n=1}^{16}{2n}$
Penyelesaian: Lihat/Tutup $2.1+2.2+2.3+...+2.16=\sum\limits_{n=1}^{16}{2n}$
Jawaban: E
Soal No. 23
Notasi sigma yang menyatakan penjumlahan beruntun 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 adalah ….
A. $\sum\limits_{n=1}^{8}{(4n-3)}$
B. $\sum\limits_{n=1}^{8}{(3n-2)}$
C. $\sum\limits_{n=2}^{9}{(4n-3)}$
D. $\sum\limits_{n=2}^{9}{(3n-2)}$
E. $\sum\limits_{n=0}^{7}{(3n+3)}$
Penyelesaian: Lihat/Tutup 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22
= (3.1 – 2) + (3.2 – 2) + (3.3 – 2) + … + (3.8 – 2)
= $\sum\limits_{n=1}^{8}{(3n-2)}$
Jawaban: B
Soal No. 24
Nilai dari $\sum\limits_{i=1}^{10}{(i+2)}$ = ….
A. 63
B. 65
C. 73
D. 75
E. 77
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{10}{(i+2)} &= \sum\limits_{i=1}^{10}{i}+\sum\limits_{i=1}^{10}{2} \\ &= \frac{10(10+1)}{2}+10\times 2 \\ &= 55+20 \\ &= 75 \end{align}$
Jawaban: D
Soal No. 25
Nilai dari $\sum\limits_{i=3}^{8}{(4i-6)}$ = ….
A. 108
B. 102
C. 96
D. 88
E. 72
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=3}^{8}{(4i-6)} &= \sum\limits_{i=3-2}^{8-2}{(4(i+2)-6)} \\ &= \sum\limits_{i=1}^{6}{(4i+8-6)} \\ &= \sum\limits_{i=1}^{6}{(4i+2)} \\ &= 4\sum\limits_{i=1}^{6}{i}+\sum\limits_{i=1}^{6}{2} \\ &= 4.\frac{6(6+1)}{2}+6\times 2 \\ &= 96 \end{align}$
Jawaban: C
Soal No. 26
Nilai dari $\sum\limits_{k=1}^{5}{\frac{k-2}{k}}$ adalah ….
A. $\frac{1}{2}$
B. $\frac{13}{30}$
C. $\frac{2}{5}$
D. $\frac{11}{30}$
E. $\frac{1}{3}$
Penyelesaian: Lihat/Tutup $\sum\limits_{k=1}^{5}{\frac{k-2}{k}}$
= $\frac{1-2}{1}+\frac{2-2}{2}+\frac{3-2}{3}+\frac{4-2}{4}+\frac{5-2}{5}$
= $-1+0+\frac{1}{3}+\frac{1}{2}+\frac{3}{5}$
= $\frac{-1\times 30}{30}+\frac{1\times 10}{3\times 10}+\frac{1\times 15}{2\times 15}+\frac{3\times 6}{5\times 6}$
= $\frac{-30+10+15+18}{30}$
= $\frac{13}{30}$
Jawaban: B
Soal No. 27
Hasil dari $\sum\limits_{n=1}^{7}{3n}+\sum\limits_{n=1}^{7}{(n-1)}$ = ….
A. 84
B. 95
C. 102
D. 104
E. 105
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=1}^{7}{3n}+\sum\limits_{n=1}^{7}{(n-1)} &= \sum\limits_{n=1}^{7}{(3n+n-1)} \\ &= \sum\limits_{n=1}^{7}{(4n-1)} \\ &= 4\sum\limits_{n=1}^{7}{n}-\sum\limits_{n=1}^{7}{1} \\ &= 4.\frac{7(7+1)}{2}-7.1 \\ &= 105 \end{align}$
Jawaban: E
Soal No. 28
ilai dari $\sum\limits_{i=1}^{9}{(3i-2)\frac{i}{6}}$ adalah ….
A. 132,0
B. 131,5
C. 130,0
D. 127,5
E. 126,0
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{9}{(3i-2)\frac{i}{6}}$
= $\sum\limits_{i=1}^{9}{\left( \frac{1}{2}{{i}^{2}}-\frac{1}{3}i \right)}$
= $\frac{1}{2}\sum\limits_{i=1}^{9}{{{i}^{2}}}-\frac{1}{3}\sum\limits_{i=1}^{9}{i}$
= $\frac{1}{2}.\frac{9(9+1)(9.2+1)}{6}-\frac{1}{3}.\frac{9(9+1)}{2}$
= 142,5 – 15
= 127,5
Jawaban: D
Soal No. 29
Diketahui $\sum\limits_{k=1}^{10}{(2+ak)}=0$, nilai $a$ = ….
A. $-\frac{11}{4}$
B. $-\frac{4}{11}$
C. $-\frac{2}{11}$
D. $\frac{4}{11}$
E. $\frac{11}{2}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{10}{(2+ak)} &= 0 \\ \sum\limits_{k=1}^{10}{2}+a\sum\limits_{k=1}^{10}{k} &= 0 \\ 10.2+a.\frac{10(10+1)}{2} &= 0 \\ 20+55a &= 0 \\ 55a &= -20 \\ a &= -\frac{20}{55} \\ a &= -\frac{4}{11} \end{align}$
Jawaban: B
Soal No. 30
Diketahui $\sum\limits_{k=1}^{n}{(2k-1)}=225$, nilai $n$ = ….
A. 10
B. 15
C. 20
D. 15
E. 30
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{n}{(2k-1)} &= 225 \\ 2\sum\limits_{k=1}^{n}{k}-\sum\limits_{k=1}^{n}{1} &= 225 \\ 2.\frac{n(n+1)}{2}-n.1 &= 225 \\ n(n+1)-n &= 225 \\ n^2+n-n &= 225 \\ n^2 &= 225 \\ n &= 15 \end{align}$
Jawaban: B
Soal No. 31
Jika batas bawah diubah menjadi sepuluh, bentuk notasi sigma $\sum\limits_{n=2}^{12}{(n^2-3n)}$ akan menjadi ….
A. $\sum\limits_{n=10}^{20}{(n^2-19n+40)}$
B. $\sum\limits_{n=10}^{20}{(n^2-19n+88)}$
C. $\sum\limits_{n=10}^{20}{(n^2-13n+40)}$
D. $\sum\limits_{n=10}^{20}{(n^2-13n+88)}$
E. $\sum\limits_{n=10}^{20}{(n^2+13n+40)}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=2}^{12}{(n^2-3n)}$
= $\sum\limits_{n=2+8}^{12+8}{({{(n-8)}^{2}}-3(n-8))}$
= $\sum\limits_{n=10}^{20}{(n^2-16n+64-3n+24)}$
= $\sum\limits_{n=10}^{20}{(n^2-19n+88)}$
Jawaban: B
Soal No. 32
Jika bentuk $\sum\limits_{n=4}^{8}{{{(n-2)}^{2}}}$ diubah dengan notasi sigma lain yang sesuai dengan batas bawah satu, akan diperoleh bentuk ….
A. $\sum\limits_{n=1}^{5}{(n^2-2n+1)}$
B. $\sum\limits_{n=1}^{5}{(n^2+2n+1)}$
C. $\sum\limits_{n=1}^{5}{(n^2-4n-4)}$
D. $\sum\limits_{n=1}^{5}{(n^2-10n+25)}$
E. $\sum\limits_{n=1}^{5}{(n^2+10n+25)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=4}^{8}{{{(n-2)}^{2}}} &= \sum\limits_{n=4-3}^{8-3}{{{((n+3)-2)}^{2}}} \\ &= \sum\limits_{n=1}^{5}{{{(n+1)}^{2}}} \\ &= \sum\limits_{n=1}^{5}{(n^2+2n+1)} \end{align}$
Jawaban: B
Soal No. 33
$\sum\limits_{i=1}^{n}{(4i-3)}$ = ….
A. $n(n+2)$
B. $n(2n-1)$
C. $n(2n+1)$
D. $n(3n-1)$
E. $n(3n+2)$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{n}{(4i-3)}$
= $4\sum\limits_{i=1}^{n}{i}-\sum\limits_{i=1}^{n}{3}$
= $4.\frac{n(n+1)}{2}-n.3$
= $2n(n+1)-3n$
= $2n^2+2n-3n$
= $2n^2-n$
= $n(2n-1)$
Jawaban: B
Soal No. 34
$\sum\limits_{k=1}^{10}{(2^k-2^{k-1})}$ = ….
A. $1-{{2}^{10}}$
B. ${{2}^{10}}-1$
C. $1+{{2}^{10}}$
D. $2-{{2}^{9}}$
E. $2+{{2}^{10}}$
Penyelesaian: Lihat/Tutup $\sum\limits_{k=1}^{10}{(2^k-2^{k-1})}$
= $\sum\limits_{k=1}^{10}{(-2^{k-1}+2^k)}$
= $(-{{2}^{0}}+{{2}^{1}})+(-{{2}^{1}}+2^2)+(-2^2+{{2}^{3}})+...+(-{{2}^{9}}+{{2}^{10}})$
= $-{{2}^{0}}+{{2}^{1}}-{{2}^{1}}+2^2-2^2+{{2}^{3}}+...-{{2}^{9}}+{{2}^{10}}$
= $-{{2}^{0}}+{{2}^{10}}$
= $-1+{{2}^{10}}$
= ${{2}^{10}}-1$
Jawaban: B
Soal No. 35
Nilai dari $\sum\limits_{n=1}^{50}{[{{(-1)}^{n}}.n]}$ = ….
A. 50
B. 25
C. 0
D. $-25$
E. $-50$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{50}{[{{(-1)}^{n}}.n]}$
= $-1+2-3+4-...-47+48-49+50$
= $\underbrace{1+1+1+...+1}_{sebanyak\,25}$
= 25
Jawaban: B
Soal No. 36
Nilai dari $\sum\limits_{n=1}^{160}{[{{(-1)}^{n}}(3n+7)]}$ = ….
A. 240
B. 80
C. 0
D. $-80$
E. $-240$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{160}{[{{(-1)}^{n}}(3n+7)]}$
= $-10+13-16+19-....-478+481-484+487$
= $\underbrace{3+3+3+...+3+3}_{sebanyak\,80}$
= $80\times 3$
= 240
Jawaban: A
Soal No. 37
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{k=1}^{n}{{{(k-3)}^{2}}}$ adalah ….
A. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-9$
B. $\sum\limits_{k=1}^{n}{{{k}^{2}}}+6\sum\limits_{k=1}^{n}{k}+9$
C. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9$
D. $\sum\limits_{k=1}^{n}{{{k}^{2}}}+6\sum\limits_{k=1}^{n}{k}+9n$
E. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9n$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{n}{{{(k-3)}^{2}}} &= \sum\limits_{k=1}^{n}{({{k}^{2}}-6k+9)} \\ &= \sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{9} \\ &= \sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9n \end{align}$
Jawaban: E
Soal No. 38
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{p=1}^{n}{({{p}^{2}}-p+1)(p+1)}$ adalah ….
A. $\sum\limits_{p=1}^{n}{{{p}^{3}}}+n$
B. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-n$
C. $\sum\limits_{p=1}^{n}{{{p}^{3}}}+2\sum\limits_{p=1}^{n}{{{p}^{2}}}+2\sum\limits_{p=1}^{n}{p}+n$
D. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-2\sum\limits_{p=1}^{n}{{{p}^{2}}}+2\sum\limits_{p=1}^{n}{p}-n$
E. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-2\sum\limits_{p=1}^{n}{{{p}^{2}}}-2\sum\limits_{p=1}^{n}{p}-n$
Penyelesaian: Lihat/Tutup $\sum\limits_{p=1}^{n}{({{p}^{2}}-p+1)(p+1)}$
= $\sum\limits_{p=1}^{n}{({{p}^{3}}+{{p}^{2}}-{{p}^{2}}-p+p+1)}$
= $\sum\limits_{p=1}^{n}{({{p}^{3}}+1)}$
= $\sum\limits_{p=1}^{n}{{{p}^{3}}}+\sum\limits_{p=1}^{n}{1}$
= $\sum\limits_{p=1}^{n}{{{p}^{3}}}+n$
Jawaban: A
Soal No. 39
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{p=4}^{18}{{{(3p-1)}^{2}}}$ adalah ….
A. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+64$
B. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+960$
C. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}-8\sum\limits_{p=1}^{15}{p}+64$
D. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}-8\sum\limits_{p=1}^{15}{p}+960$
E. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+6\sum\limits_{p=1}^{15}{p}+15$
Penyelesaian: Lihat/Tutup $\sum\limits_{p=4}^{18}{{{(3p-1)}^{2}}}$
= $\sum\limits_{p=4-3}^{18-3}{{{(3(p+3)-1)}^{2}}}$
= $\sum\limits_{p=1}^{15}{{{(3p+9-1)}^{2}}}$
= $\sum\limits_{p=1}^{15}{{{(3p+8)}^{2}}}$
= $\sum\limits_{p=1}^{15}{(9{{p}^{2}}+48p+64)}$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+\sum\limits_{p=1}^{15}{64}$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+15\times 64$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+960$
Jawaban: B
Soal No. 40
Nilai dari $\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}$ = ….
A. $\frac{45}{27}$
B. $\frac{46}{27}$
C. $\frac{47}{27}$
D. $\frac{48}{27}$
E. $\frac{49}{27}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}\,....\,(1)$
$\begin{align}S &= 7\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k}}}} \\ \frac{1}{7}S &= \sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k}}}} \\ &= \sum\limits_{k=1+1}^{\infty }{\frac{{{(k-1)}^{2}}}{{{7}^{k-1}}}} \\ \frac{1}{7}S &= \sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}S-\frac{1}{7}S &= \sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}-\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= \frac{1}{{{7}^{1-1}}}+\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}-\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= 1+\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,....\,(3) \end{align}$
$P=\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,...\,(4)$
$\begin{align}P &= 7\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k}}}} \\ \frac{1}{7}P &= \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k}}}} \\ &= \sum\limits_{k=2+1}^{\infty +1}{\frac{2(k-1)-1}{{{7}^{k-1}}}} \\ \frac{1}{7}P &= \sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}}\,....\,(5) \end{align}$
Persamaan (4) dikurangkan dengan persamaan (5) maka:
$\begin{align}P-\frac{1}{7}P &= \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,-\sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}} \\ \frac{6}{7}P &= \frac{2.2-1}{{{7}^{2-1}}}+\sum\limits_{k=3}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}-\sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}} \\ &= \frac{3}{7}+\sum\limits_{k=3}^{\infty }{\frac{2}{{{7}^{k-1}}}} \\ &= \frac{3}{7}+\left( \frac{2}{{{7}^{2}}}+\frac{2}{{{7}^{3}}}+\frac{2}{{{7}^{4}}}+.... \right) \\ &= \frac{3}{7}+\frac{\frac{2}{49}}{1-\frac{1}{7}} \\ &= \frac{3}{7}+\frac{2}{49}.\frac{7}{6} \\ &= \frac{3}{7}+\frac{1}{21} \\ \frac{6}{7}P &= \frac{10}{21} \\ P &= \frac{10}{21}\times \frac{7}{6} \\ P &= \frac{5}{3}\times \frac{1}{3} \\ \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}} &= \frac{5}{9} \end{align}$
Substitusi ke persamaan (3):
$\begin{align}\frac{6}{7}S &= 1+\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= 1+\frac{5}{9} \\ \frac{6}{7}S &= \frac{14}{9} \\ S &= \frac{14}{9}\times \frac{7}{6} \\ S &= \frac{49}{27} \end{align}$
Jawaban: E
Soal No. 41
$\sum\limits_{i=1}^{15}{\frac{4}{(2n+3)(2n+1)}}$ = ….
A. $\frac{17}{33}$
B. $\frac{19}{33}$
C. $\frac{20}{33}$
D. $\frac{23}{33}$
E. $\frac{25}{33}$
Penyelesaian: Lihat/Tutup $\frac{4}{(2n+3)(2n+1)}=\frac{a}{2n+3}+\frac{b}{2n+1}$
$\frac{4}{(2n+3)(2n+1)}=\frac{a(2n+1)+b(2n+3)}{(2n+3)(2n+1)}$
$4=a(2n+1)+b(2n+3)$
$n=-\frac{1}{2}$ maka:
$4=a(2n+1)+b(2n+3)$
$4=a\left( 2\left( -\frac{1}{2} \right)+1 \right)+b\left( 2\left( -\frac{1}{2} \right)+3 \right)$
$4=a\left( -1+1 \right)+b\left( -1+3 \right)$
$4=2b$
$b=2$
$n=-\frac{3}{2}$ maka:
$4=a(2n+1)+b(2n+3)$
$4=a\left( 2\left( -\frac{3}{2} \right)+1 \right)+b\left( 2\left( -\frac{3}{2} \right)+3 \right)$
$4=a\left( -3+1 \right)+b\left( -3+3 \right)$
$4=-2a$
$a=-2$
Jadi,
$\sum\limits_{i=1}^{15}{\frac{4}{(2n+3)(2n+1)}}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{a}{(2n+3)}+\frac{b}{(2n+1)} \right]}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{-2}{(2n+3)}+\frac{2}{(2n+1)} \right]}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{2}{(2n+1)}-\frac{2}{(2n+3)} \right]}$
= $\left( \frac{2}{3}-\frac{2}{5} \right)$ + $\left( \frac{2}{5}-\frac{2}{7} \right)$ + $\left( \frac{2}{7}-\frac{2}{9} \right)$ + … + $\left( \frac{2}{29}-\frac{2}{31} \right)$ + $\left( \frac{2}{31}-\frac{2}{33} \right)$
= $\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}$ + … + $\frac{2}{29}-\frac{2}{31}+\frac{2}{31}-\frac{2}{33}$
= $\frac{2}{3}-\frac{2}{33}$
= $\frac{22-2}{33}$
= $\frac{20}{33}$
Jawaban: C
Soal No. 42
$\sum\limits_{k=1}^{\infty }{\left( \frac{1}{(3k+1)(3k+4)} \right)}$ = ….
A. $\frac{1}{13}$
B. $\frac{1}{12}$
C. $\frac{1}{11}$
D. $\frac{1}{10}$
E. $\frac{1}{9}$
Penyelesaian: Lihat/Tutup $\begin{align}\frac{1}{(3k+1)(3k+4)} &= \frac{a}{3k+1}+\frac{b}{3k+4} \\ \frac{1}{(3k+1)(3k+4)} &= \frac{a(3k+4)+b(3k+1)}{(3k+1)(3k+4)} \\ 1 &= a(3k+4)+b(3k+1) \end{align}$
$k=-\frac{1}{3}$ maka:
$\begin{align}1 &= a(3k+4)+b(3k+1) \\ 1 &= 3a \\ a &= \frac{1}{3} \end{align}$
$k=-\frac{4}{3}$ maka:
$\begin{align}1 &= a(3k+4)+b(3k+1) \\ 1 &= -3b \\ b &= -\frac{1}{3} \end{align}$
$\sum\limits_{k=1}^{\infty }{\left( \frac{1}{(3k+1)(3k+4)} \right)}$
= $\sum\limits_{k=1}^{\infty }{\left( \frac{1/3}{3k+1}-\frac{1/3}{3k+4} \right)}$
= $\frac{1}{3}\sum\limits_{k=1}^{\infty }{\left( \frac{1}{3k+1}-\frac{1}{3k+4} \right)}$
= $\frac{1}{3}\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \frac{1}{3k+1}-\frac{1}{3k+4} \right)}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{3n+1}-\frac{1}{3n+4} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{1}{4}-\frac{1}{3n+4} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{3n+4-4}{4(3n+4)} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{12n+16}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{n}{n}}{\frac{12n}{n}+\frac{16}{n}}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{12+\frac{16}{n}}$
= $\frac{1}{12+\frac{16}{\infty }}$
= $\frac{1}{12+0}$
= $\frac{1}{12}$
Jawaban: B
Soal No. 43
Bentuk aljabar $x_1{{y}^{8}}+x_2{{y}^{9}}+{{x}_{3}}{{y}^{10}}+...+{{x}_{16}}{{y}^{23}}$ dapat dinyatakan dalam bentuk notasi sigma, yaitu ….
A. $\sum\limits_{n=0}^{16}{{{x}_{n}}.{{y}^{n+7}}}$
B. $\sum\limits_{n=1}^{8}{{{x}_{2n-1}}.{{y}^{2n+5}}}$
C. $\sum\limits_{n=3}^{18}{{{x}_{n-2}}.{{y}^{n+5}}}$
D. $\sum\limits_{n=2}^{17}{{{x}_{n-1}}.{{y}^{n+5}}}$
E. $\sum\limits_{n=-3}^{13}{{{x}_{n+3}}.{{y}^{n+8}}}$
Penyelesaian: Lihat/Tutup $x_1{{y}^{8}}+x_2{{y}^{9}}+{{x}_{3}}{{y}^{10}}+...+{{x}_{16}}{{y}^{23}}$
= $x_1{{y}^{1+7}}+x_2{{y}^{2+7}}+{{x}_{3}}{{y}^{3+7}}+...+{{x}_{16}}{{y}^{16+7}}$
= $\sum\limits_{n=1}^{16}{{{x}_{n}}{{y}^{n+7}}}$
= $\sum\limits_{n=1+2}^{16+2}{{{x}_{(n-2)}}{{y}^{(n-2)+7}}}$
= $\sum\limits_{n=3}^{18}{{{x}_{n-2}}.{{y}^{n+5}}}$
Jawaban: C
Soal No. 44
Nilai dari $\sum\limits_{n=1}^{70}{[{{(-1)}^{n}}(2n-5)]}$ = ….
A. 70
B. 35
C. 0
D. $-35$
E. $-70$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{70}{[{{(-1)}^{n}}(2n-5)]}$
= $-(-3)+(-1)-1+3-...-129+131-133+135$
= $\underbrace{2+2+...+2+2}_{sebanyak\,35}$
= $35\times 2$
= 70
Jawaban: A
Soal No. 45
Nilai dari $\sum\limits_{i=1}^{20}{\left( \frac{1}{i}-\frac{1}{i+1} \right)}$ adalah ….
A. $\frac{1}{21}$
B. $\frac{1}{20}$
C. $\frac{9}{21}$
D. $\frac{19}{20}$
E. $\frac{20}{21}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{20}{\left( \frac{1}{i}-\frac{1}{i+1} \right)}$
= $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}+\frac{1}{20}-\frac{1}{21}$
= $1-\frac{1}{21}$
= $\frac{20}{21}$
Jawaban: E
Soal No. 46
Nilai dari $\sum\limits_{n=1}^{10}{({{2}^{n}}-{{2}^{n-1}})}$ adalah ….
A. 1.025
B. 1.024
C. 1.023
D. $-1.023$
E. $-1.025$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{10}{({{2}^{n}}-{{2}^{n-1}})}$
= $\sum\limits_{n=1}^{10}{(-{{2}^{n-1}}+{{2}^{n}})}$
= $-{{2}^{0}}+{{2}^{1}}-{{2}^{1}}+2^2-2^2+{{2}^{3}}-...-{{2}^{8}}+{{2}^{9}}-{{2}^{9}}+{{2}^{10}}$
= $-{{2}^{0}}+{{2}^{10}}$
= $-1+1024$
= 1023
Jawaban: C
Soal No. 47
Nilai dari $\sum\limits_{i=1}^{38}{\left( \frac{i}{i+1}-\frac{i+1}{i+2} \right)}$ adalah ….
A. $1\frac{19}{40}$
B. $\frac{19}{40}$
C. 0
D. $-\frac{19}{40}$
E. $-1\frac{19}{40}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{38}{\left( \frac{i}{i+1}-\frac{i+1}{i+2} \right)}$
= $\frac{1}{2}-\frac{2}{3}+\frac{2}{3}-\frac{3}{4}+\frac{3}{4}-\frac{4}{5}+...+\frac{37}{38}-\frac{38}{39}+\frac{38}{39}-\frac{39}{40}$
= $\frac{1}{2}-\frac{39}{40}$
= $-\frac{19}{40}$
Jawaban: D
Soal No. 48
Bentuk penjumlahan beruntun dari notasi sigma $\sum\limits_{n=1}^{4}{\left( \frac{1}{n+1}+\frac{1}{n} \right)}$ adalah ….
A. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$
B. $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$
C. $\frac{1}{2}+\frac{5}{6}+\frac{1}{4}+\frac{9}{20}$
D. $\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{5}{6}$
E. $\frac{3}{2}+\frac{5}{6}+\frac{7}{12}+\frac{9}{20}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{4}{\left( \frac{1}{n+1}+\frac{1}{n} \right)}$
= $\sum\limits_{n=1}^{4}{\left( \frac{n}{n(n+1)}+\frac{n+1}{n(n+1)} \right)}$
= $\sum\limits_{n=1}^{4}{\left( \frac{2n+1}{n(n+1)} \right)}$
= $\frac{3}{2}+\frac{5}{6}+\frac{7}{12}+\frac{9}{20}$
Jawaban: E
Soal No. 49
Bentuk penjumlahan beruntun dari notasi sigma $\sum\limits_{n=1}^{5}{\left[ {{(-1)}^{n}}.\frac{n+1}{2n+1} \right]}$ adalah ….
A. $\frac{2}{3}+\frac{3}{5}+\frac{4}{7}+\frac{5}{9}+\frac{6}{11}$
B. $-\frac{2}{3}+\frac{3}{5}+\frac{4}{7}+\frac{5}{9}+\frac{6}{11}$
C. $-\frac{2}{3}+\frac{3}{5}-\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
D. $-\frac{2}{3}-\frac{3}{5}+\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
E. $-\frac{2}{3}-\frac{3}{5}-\frac{4}{7}-\frac{5}{9}-\frac{6}{11}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{5}{\left[ {{(-1)}^{n}}.\frac{n+1}{2n+1} \right]}=-\frac{2}{3}+\frac{3}{5}-\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
Jawaban: C
Soal No. 50
Notasi sigma yang menyatakan penjumlahan beruntun $2-4+8-16+32-64$ adalah ….
A. $-2\sum\limits_{n=1}^{6}{{{2}^{n-1}}}$
B. $-2\sum\limits_{n=1}^{6}{{{2}^{n+1}}}$
C. $-2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
D. $2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
E. $2\sum\limits_{n=1}^{6}{{{(-2)}^{n+1}}}$
Penyelesaian: Lihat/Tutup $2-4+8-16+32-64$
= ${{2.2}^{0}}-{{2.2}^{1}}+{{2.2}^{2}}-{{2.2}^{3}}+{{2.2}^{4}}-{{2.2}^{5}}$
= $2{{(-2)}^{0}}+2{{(-2)}^{1}}+2{{(-2)}^{2}}+2{{(-2)}^{3}}+2{{(-2)}^{4}}+2{{(-2)}^{5}}$
= $\sum\limits_{n=1}^{6}{2{{(-2)}^{n-1}}}$
= $2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
Jawaban: D
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