Question

The sum of the series $1+2.2+3.2^2+4.2^3+5.2^4+..........+100.2^{99}$ is

• A. $99.2^{100}$
• B. $99.2^{100}+1$
• C. $100.2^{100}$
• D. none of these

Answer 1

Answer: (B)

Let $S=1+2.2+3.2^2+4.2^3+......+100.2^{99}$
$\therefore 2S=1.2+2.2^2+3.2^3+......+99.2^{99}+100.2^{100}$
Subtracting, we get
$-S=1+1.2+1.2^2+1.2^3+.....+1.2^{99}-100.2^{100}$
$=\left(1+2+2^2+......2^{99}\right)-100.2^{100}$
$=\frac{1\left(2^{100}-1\right)}{2-1}-100.2^{100}=2^{100}-1-100.2^{100}$
$=\frac{1\left(2^{100}-1\right)}{2-1}-100.2^{100}$
$=2^{100}-1-100.2^{100}$
$\therefore S=100.2^{100}-2^{100}+1$
$=99.2^{100}+1$