Answer to Question #223287 in Calculus for Rasgel

Question

Simplify the sum Σ from r=0 to 2n+1 of [3/(r+1)(r+2)]

Answer 1

$\frac{3}{(r+1)(r+2)}=\frac{3}{r+1}-\frac{3}{r+2}.$

$\Sigma_{r=0}^{2n+1}\frac{3}{(r+1)(r+2)}=\Sigma_{r=0}^{2n+1}[\frac{3}{r+1}-\frac{3}{r+2}]=$

$=\frac{3}{1}-\frac{3}{2n+1+2}=3-\frac{3}{2n+3}=\frac{6(n+1)}{2n+3}.$