Question

What is the sum `1/2+1/6+1/12+1/20+...+1/(a^2-a)` ?

Answer 1

`1/2+1/6+1/12+1/20+...+1/(a^2-a)=1-1/a`

Explanation:

`1/2+1/6+1/12+1/20+...+1/(a^2-a)`

`=sum_(k=2)^a 1/(k^2-k)`

`=sum_(k=2)^a 1/(k(k-1))`

`=sum_(k=2)^a (1/(k-1)-1/k)`

`=sum_(k=2)^a 1/(k-1)- sum_(k=2)^a 1/k`

`=1+color(red)(cancel(color(black)(sum_(k=2)^(a-1) 1/k))) - color(red)(cancel(color(black)(sum_(k=2)^(a-1) 1/k))) - 1/a`

`=1-1/a`