Question

Find the sum of $n$ terms of the series the $r^{th}$ term of which is $(2r+1)2^r$.

Answer 1

$(2r+1)2^r=r{2}^{r+1}+2^r$
Let $S_1=\sum _{r=1}^n{2}^r=\frac{2(2^n-1)}{2-1}=2^{n+1}-2$
$S_2=\sum _{r=1}^{n}r{2}^{r+1}={1.2}^2+{2.2}^3+.......n{.2}^{n+1}$ ..... $\left(i\right)$
Multiply both sides by $2$, we get
$2S_2=\sum _{r=1}^{n}r{2}^{r+1}=0+{1.2}^3+{2.2}^4+......(n-1){.2}^{n+1}+n{.2}^{n+2}$ ..... $\left(ii\right)$
Subtracting both equations $\left(i\right)$ and $\left(ii\right)$, we get
$-S_2=\sum _{r=1}^{n}r{2}^{r+1}=2^2+2^3+...........{.2}^{n+1}-n{.2}^{n+2}$
$-S_2=\frac{4[2^n-1]}{2-1}-n{2}^{n+2}$
$S_2=-\left(4\right[2^n-1\left]\right)+n{2}^{n+2}$
Total sum = $S_1+S_2=2^{n+1}-2-4[2^n-1]+n{2}^{n+2}=n{.2}^{n+2}-2^{n+1}+2$