Question

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

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

Answer 1

The correct option is D ${99.2}^{100}+1$

$S=1+2.2+{3.2}^2+{4.2}^3+{5.2}^4+\cdots +{100.2}^{99}\therefore 2S=1.2+{2.2}^2+{3.2}^3+\cdots +{99.2}^{99}+{100.2}^{100}$
Subtracting we get
$-S=1+1.2+{1.2}^2+\cdots +{1.2}^{99}-{100.2}^{100}=(1+2+2^2+\cdots +2^{99})-{100.2}^{100}=\frac{1(2^{100}-1)}{2-1}-{100.2}^{100}=2^{100}-1-{100.2}^{100}\therefore S={100.2}^{100}-2^{100}+1={99.2}^{100}+1$