The sum to 50 terms of the series
`3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is `

Options

A. `(6n)/(n+1)`
B. `(9n)/(n+1)`
C. `(12n)/(n+1)`
D. `(3n)/(n+1)`

Answer

A

Explanation

Let `T_(r)` be the `r^(th)` term of the given series. Then,
`T_(r)=(2r+1)/(1^(2)+2^(2)+ . . . +r^(2))=(2r+1)/((r//6)(r+1)(2r+1))=6((1)/(r)-(1)/(r+1))`
So, required sum is given by
`underset(r=1)overset(n)sumT_(r)=6underset(r=1)overset(n)sum((1)/(r)-(1)/(r+1))`
`rArr" "underset(r=1)overset(n)sum*T_(r)=6{(1-(1)/(2))+((1)/(2)-(1)/(3))+((1)/(3)-(1)/(4))+ . . . +((1)/(n)-(1)/(n+1))}`
`rArr" "underset(r=1)overset(n)sumT_(r)=6{1-(1)/(n+1)}=(6n)/(n+1)`