Question

`1/(1xx2)+1/(4xx7)+1/(2xx3)+1/(7xx10)+...` to `20` terms `=`?

Answer 1

`735/748`

Explanation:

The required value can be found by adding two sums

`S_1 = 1/{1times 2}+1/{2 times 3}+...+1/{10 times 11}`

and

`S_2 = 1/{4times 7}+1/{7times 10}+...+1/{31 times 34}`

Now, since `1/{n(n+a) } = 1/a(1/n-1/{n+a})`, we have

`S_1 = 1/1-1/2+1/2-1/3+...+1/10-1/11 = 1/1-1/11 = 10/11`
and
`S_2 = 1/3(1/4-1/7+1/7-1/10+...+1/31-1/34)=1/3(1/4-1/34) = {17-2}/{3times68}=5/68`

Thus, the required sum is
`10/11+5/68 = {10 times 68+5times 11}/{11times 68}=735/748`