sum upto n terms where rth term is $r(r+1)2^r$

Question

sum upto n terms where rth term is $r(r+1)2^r$.

I tried to make a telescoping series but failed.It seems like i have to subtract and add something from r(r+1) such that power of 2 also change.Is there a systematic approach?

Answer 1

Hint:

$$S=\sum_{r=1}^n r(r+1)x^r$$

$$(1-x)S=2\sum_{r=1}^nrx^r-n(n+1)x^{n+1}\text{ as } r(r+1)-r(r-1)=2r$$

Again, if $$T=\sum_{r=1}^nrx^r$$

$$(1-x)T=?$$

Answer 2

Take the Ansatz $\sum_{r=1}^nr(r+1)2^r=(an^2+bn+c)2^n-c$ so the $n=0$ case works, while the inductive step reduces to$$\begin{align}n(n+1)2^n&=(an^2+bn+c)2^n-(a(n-1)^2+b(n-1)+c)2^{n-1}\\&=(an^2-a(n-1)^2/2+bn-b(n-1)/2)2^n\\&=(an^2/2+(a+b/2)n+(b-a+c)/2)2^n.\end{align}$$So $a=2,\,b=-2,\,c=4$.

Answer 3

From $$r(r+1)-2(r-1)r+(r-2)(r-1)=2$$

you can deduce

$$S_n-4S_{n-1}+4S_{n-2} \\=\sum_{r=1}^nr(r+1)2^r -2\cdot2\sum_{r=1}^n(r-1)r2^{r-1} +2^2\sum_{r=1}^n (r-2)(r-1)2^{r-2} \\=2\sum_{r=1}^n 2^r,$$

which is

$$S_n-4(S_n-n(n+1)2^n)+4(S_n-(n-1)n2^{n-1})=2\frac{2^{n+1}-1}2.$$

You can draw $S_n$.