Question

SUM the following series, with every other successive denominator in the bracket being greater than the previous one by "1" ?

`1/3[(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+....]`

Answer 1

`13/36`

Explanation:

first of all, divide the sum into to parts- the positive constants and the negative constants.
`=1/3{[1/2+1/3+1/4+1/5+......]-[1/5+1/6+1/7+......]}`
cancel the constants of opposite signs,i.e., from `1/5` to infinity.
`=1/3{1/2+1/3+1/4}`
take LCM on the inside
`=1/3{[6+4+3]/12}`
`=13/36`

hope this helps,
Shivang M.