SOLUTION: [n(sigma)r=1]{{{1/(r(r+2))}}}=[n(sigma)r=1]{{{(1/2)((1/r)-(1/(r+2)))}}} ={{{(1/2)((1/1)+(1/2)-(1/(n+1))-(1/(n+2))))}}} Please show me more detailed steps, especially between the

Algebra ->  Sequences-and-series -> SOLUTION: [n(sigma)r=1]{{{1/(r(r+2))}}}=[n(sigma)r=1]{{{(1/2)((1/r)-(1/(r+2)))}}} ={{{(1/2)((1/1)+(1/2)-(1/(n+1))-(1/(n+2))))}}} Please show me more detailed steps, especially between the       Log On


   



Question 1035284: [n(sigma)r=1]1%2F%28r%28r%2B2%29%29=[n(sigma)r=1]%281%2F2%29%28%281%2Fr%29-%281%2F%28r%2B2%29%29%29
=%281%2F2%29%28%281%2F1%29%2B%281%2F2%29-%281%2F%28n%2B1%29%29-%281%2F%28n%2B2%29%29%29%29
Please show me more detailed steps, especially between the last two steps.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
This is what you call a telescoping series.

= 1%2F2(1-1%2F3+%2B1%2F2+-+1%2F4+%2B+1%2F3+-+1%2F5+...+1%2F%28n-2%29-1%2Fn+%2B+1%2F%28n-1%29+-+1%2F%28n%2B1%29+%2B+1%2Fn+-+1%2F%28n%2B2%29)
=%281%2F2%29%281+%2B+1%2F2+-1%2F%28n%2B1%29+-+1%2F%28n%2B2%29%29
= 3%2F4+-+%282n%2B3%29%2F%282n%5E2%2B6n%2B4%29