First slide

Special Series in Sequences and Series

Question

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

NA

A

4/2
B

9/2
C

4/9
D

1

Solution

This is clearly not an arithmetico-geometric series,
since 1, 4, 9, '16, ... are not in AP. However, their successive differences (4 - 1), (9 - 4), (16 - 9) ,... are in AP.
Let
$\begin{matrix} S_{\infty} = 1 + \frac{4}{3} + \frac{9}{3^{2}} + \frac{16}{3^{3}} + \dots \infty \\ \frac{1}{3} S_{\infty} = \frac{1}{3} + \frac{4}{3^{2}} + \frac{9}{3^{3}} + \dots \infty \end{matrix}$
On subtraction, we get
$\begin{aligned} \frac{2}{3} S_{\infty} = 1 + \frac{3}{3} + \frac{5}{3^{2}} + \frac{7}{3^{3}} + \dots \infty \\ \Rightarrow \frac{1}{3} \cdot \frac{2}{3} S_{\infty} = \frac{1}{3} + \frac{3}{3^{2}} + \frac{5}{3^{3}} + \dots \infty \\ [multiplying both sides by \frac{1}{3}] \end{aligned}$
On subtracting the two series, we get
$\begin{array}{l} (\frac{4}{9}) \cdot S_{\infty} = 1 + \frac{2}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}} + \dots \infty \\ = 1 + \frac{2}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots \infty) = 1 + \frac{2}{3} \times \frac{1}{(1 - \frac{1}{3})} = 2 \\ ∴ S_{\infty} = (2 \times \frac{9}{4}) = \frac{9}{2} \end{array}$

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