Question

The sum of the numbers $1+2.2+3.2^2+4.2^3+...+50.2^{49}$ is

• A. $1+49·2^{49}$
• B. $1+49.2^{50}$
• C. $1+50.2^{49}$
• D. $1+50.2^{50}$

Answer 1

The correct option is B

$1+49.2^{50}$

Find the sum of the given series

$1+2.2+3.2^2+4.2^3+...+50.2^{49}$

Let the sum be $S$

$S=1+2.2+3.2^2+4.2^3+...+50.2^{49}......\left(1\right)$

Multiply by $2$

$2S=2+2·2^2+3·2^3+4·2^4........+50·2^{50}.....\left(2\right)$

Now subtracting $\left(1\right)$ from $\left(2\right)$ we get

$$\begin{gathered} 2S-S=1-(2+2^2+2^3+........2^{49})+50·2^{50} \\ \Rightarrow S=1-\left(\frac{2(2^{49}-1)}{2-1}\right)+50·2^{50} \\ \Rightarrow S=1+49·2^{50} \end{gathered}$$

Hence, the correct answer is $\mathrm{Option}\left(\mathrm{B}\right)$.