Question

Sum of the series $1+2.2+{3.2}^2+....+100\times 2^{99}$.

Answer 1

$n^{th}$ term of sum can be written as, $T_n=n\left(2^{n-1}\right)$

$\Rightarrow$ This is an A.G.P.

$S_n={1.2}^o+{2.2}^1+{3.2}^2+...\dots \dots +{100.2}^{99}$

$-2S_n={1.2}^1+{2.2}^2+...\dots ..+-{99.2}^{99}+{100.2}^{100}$

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$-S_n=1+2^1+2^2+...\dots \dots \dots +2^{99}-{100.2}^{100}$

$\Rightarrow {100.2}^{100}-S_n=(2^o+2^1+...\dots ..+2^{99})$

$=\frac{2^{100}-1}{2-1}$ $\left\{a\left(\frac{r^n-1}{r-1}\right)\right\}$

$\Rightarrow {100.2}^{100}-S_n=2^{100}-1$

$\Rightarrow S_n={100.2}^{100}-2^{100}H$

$S_n={99.2}^{100}+1$