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

2 Answers
Jan 24, 2017

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

Jan 25, 2017

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))