Question

How do you find a formula for `sum_(r=0)^n r2^r` ?

Answer 1

`sum_(r=0)^n r2^r = (n-1)2^(n+1)+2`

Explanation:

Note that:

`(n+1)2^(n+1)+sum_(r=0)^n r2^r = sum_(r=0)^(n+1) r2^r`

`color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) r2^r`

`color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) 2^r + sum_(r=1)^(n+1) (r-1)2^r`

`color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=1)^(n+1) (r-1)2^(r-1)`

`color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=0)^n r2^r`

Subtract `sum_(r=0)^n r2^r+(2^(n+2)-2)` from both ends to get:

`sum_(r=0)^n r2^r = (n+1)2^(n+1)-2^(n+2)+2`

`color(white)(sum_(r=0)^n r2^r) = (n-1)2^(n+1)+2`

Answer 2

`2(1 + 2^n (n-1))`

Explanation:

`sum_(r=0)^nrx^r=xsum_(n=0)^nrx^(r-1)=x d/(dx)(sum_(n=0)^nx^r)=`
`x d/(dx)((x^(n+1)-1)/(x-1))=(x(1 + (n (x-1)-1) x^(n)))/(x-1)^2`

now making `x=2` we have

`sum_(r=0)^nr2^r=2(1 + 2^n (n-1))`