First slide

Special Series in Sequences and Series

Question

Find the sum of the infinite series $1 + \frac{4}{3} + \frac{9}{3^{2}} + \frac{16}{3^{3}} + \dots \dots . . \infty$

Moderate

A

$\frac{7}{2}$
B

$\frac{11}{2}$
C

$\frac{13}{2}$
D

$\frac{9}{2}$

Solution

This is clearly not an AGP series, since 1, 4,9, 16…
are not in A.P. However their successive difference 4-1=3, 9-4=5, 16
9=7,… are in A.P.
$\begin{array}{l} Let S_{\infty} = 1 + \frac{4}{3} + \frac{9}{3^{2}} + \frac{16}{3^{3}} + \dots . . \infty (1) \\ \frac{1}{3} S_{\infty} = \frac{1}{3} + \frac{4}{3^{2}} + \frac{9}{3^{3}} + \dots \infty - (2) \\ Substracting (2) from (1) \\ \frac{2}{3} S_{\infty} = 1 + \frac{3}{3} + \frac{5}{3^{2}} + \frac{7}{3^{3}} + \dots . + \infty \\ \frac{1}{3} \cdot \frac{2}{3} S_{\infty} = \frac{1}{3} + \frac{3}{3^{2}} + \frac{5}{3^{3}} + \dots . . + \infty \end{array}$
$\begin{array}{l} On subtracting (\frac{4}{9}) S_{\infty} = 1 + \frac{2}{3} + \frac{2}{3^{2}} + \dots . + \infty \\ = 1 + \frac{2}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots . . + \infty) \\ = 1 + \frac{2}{3} (1 - \frac{1}{1 - \frac{1}{3}}) = 2 ∴ S_{\infty} = (2 \times \frac{9}{4}) = \frac{9}{2} \end{array}$

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