Question

The sum of the series $1+2.2+3.2^2+4.2^4+5.2^4+...+100.2^{99}$ is

• A. $99.2^{99}+1$
• B. $100.2^{100}$
• C. $99.2^{100}$
• D. $99.2^{100}+1$

Answer 1

Answer: (D)

$99.2^{100}+1$

Let $S=1+2.2+3.2^2+4.2^4+5.2^4+...+100.2^{99}$

$\therefore 2S=1.2+2.2^2+3.2^3+...+99.2^{99}+100.2^{100}$

Subtracting, we get

$-S=1+1.2+1.2^2+...+1.2^{99}-100.2^{100}$
$=(1+2+2^2+...+2^{99})-100.2^{100}$

$=\frac{1(2^{100}-1)}{2-1}-100.2^{100}=2^{100}-1-100.2^{100}$
$\therefore S=100.2^{100}-2^{100}+1=99.2^{100}+1$