Answer to Question #263099 in Algebra for Aashish

Question

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

Answer 1

Decomposing into two fractions

$\frac{3}{(r+1)(r+2)}=\frac{A}{r+1}+\frac{B}{r+2}$

Multiplying both sides by (r+1)(r+2)

$3=A(r+2)+B(r+1)$

Let $r=-2$

$\implies3=B(-2+1)=-B\implies \>B=-3$

Let $r=-1$

$\implies3=A(-1+2)\implies\>A=3$

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

Writing the first three and the last there terms

$\displaystyle\sum_{r=0}^{2n+1}(\frac {3}{r+1}-\frac{3}{r+2})=({3-\frac {3}{2}})+(\frac{3}{2}-\frac{3}{3})+(\frac{3}{3}-\frac{3}{4})+...\\\>+(\frac{3}{2n}-\>\>\frac{3}{2n+1})+(\frac{3}{2n+1}-\frac{3}{2n+2})+(\frac{3}{2n+2}-\frac{3}{2n+3})$

All fraction cancels apart from 1^st and last

$\implies$ Summation $=3-\frac{3}{2n+3}=\frac{6n+9-3}{2n+3}$

$\therefore$ $\displaystyle\sum_{r=0}^{2n+1}$ $\frac{3}{(r+1)(r+2)}=\frac{6n+6}{2n+3}=\frac{6(n+1)}{2n+3}$