Question

Find the sum of $n$ terms of the series. The rth term of which is $(2x+1)2^r.$

• A. $n{.2}^{n+3}-2^n+2.$
• B. $n{.2}^{n+3}-2^{n+1}+1.$
• C. $n{.2}^{n+2}-2^n+1.$
• D. $n{.2}^{n+2}-2^{n+1}+2.$

Answer 1

The correct option is D $n{.2}^{n+2}-2^{n+1}+2.$

$T_r=(2r+2)2^r$

$\therefore S_n=\sum _{r=1}^n{T}_r=\sum _{r=1}^n(2r+1)2^r=\sum _{r=1}^{n}2.r{2}^r+\sum _{r=1}^n{2}^r$

$=\sum _{r=1}^{n}r.2^{r+1}+\sum _{r=1}^n{2}^r$

$=(1.2^2+2.2^3+3.2^4+...+n.2^{n+1})+(2+2^2+2^3+2^4+...+2^n)$ ...(1)

Let $S=1.2^2+2.2^3+3.2^4+...+n.2^{n+1}$ ...(2)

Multiply $S$ by $2$, we get

$2S=1.2^3+2.2^4+3.2^5+...+n.2^{n+2}$ ...(3)

Applying (2)-(3), we get

$-S=1.2^2+1.2^3+1.2^4+...+1.2^{n+1}-n.2^{n+2}=2^2+2^3+2^4+...+2^{n+1}-n.2^{n+2}\Rightarrow S=n.2^{n+2}-2^2-2^3-2^4-...-2^{n+1}$

Substituting this in (1), we get

$S_n=n.2^{n+2}-2^{n+1}+2$