Question

The sum of the series 1 + 2.2 + 3.2² + 4.2³ + 5.2⁴ + .... + 100.2⁹⁹ is
(a) 99 × 2¹⁰⁰
(b) 99 × 2¹⁰⁰ + 1
(c) 100 × 2¹⁰⁰
(d) none of these

Answer 1

(b) 99 × 2¹⁰⁰ + 1

Let $S_n$ be the sum of 100 terms of the given series.

Thus, we have:

$$\begin{gathered} S_n=1+2·2+3·2^2+4·2^3+...+100·2^{99}...\left(1\right) \\ \Rightarrow 2S_n=2+2·2^2+3·2^3+...+99·2^{99}+100·2^{100}...\left(2\right) \end{gathered}$$

On subtracting (1) from (2), we get:

$$\begin{gathered} S_n=-1+\left[2·\left(-1\right)+\left(-1\right)·2^2+...+\left(-1\right)·2^{99}\right]+100·2^{100} \\ \Rightarrow S_n=-1-\left[2+2^2+...+2^{99}\right]+100·2^{100} \\ \Rightarrow S_n=-1-\left[\frac{2·\left(2^{99}-1\right)}{2-1}\right]+100·2^{100} \\ \Rightarrow S_n=-1-2^{100}+2+100·2^{100} \\ \Rightarrow S_n=1+2^{100}\left(99\right) \\ \Rightarrow S_n=99\times 2^{100}+1 \end{gathered}$$