let `S_(n)` denote the sum of the cubes of the first n natureal numbers and `S_(n)` denote the sum of the fisrt n natural numbers , then `sum_(r=1)^(n)(S_(r))/(S_(4))` equals to

Options

A. `underset(r=1)overset(n)sumr`
B. `(1)/(3)underset(r=1)overset(n+1)sumr`
C. `((n+2)/(3))underset(r=1)overset(n)sumr`
D. none of these

Answer

C

Explanation

We have,
`S_(r)=(s_(r))^(2)`
`rArr" "(S_(r))/(s_(r))=s_(r)`
`rArr" "underset(r=1)overset(n)sum(S_(r))/(s_(r))=underset(r=1)overset(n)sums_(r)underset(r=1)overset(n)sum(r(r+1))/(2)`
`underset(r=1)overset(n)sum(S_(r))/(s_(r))=(1)/(2){underset(r=1)overset(n)sumr^(2)+underset(r=1)overset(n)sumr}`
`rArr" "underset(r=1)overset(n)sum(S_(r))/(s_(r))=(1)/(2){(n(n+1)(2n+1))/(6)+(n(n+1))/(2)}`
`underset(r=1)overset(n)sum(S_(r))/(s_(r))=(n(n+1)(n+2))/(6)=((n+2)/(3))underset(r=1)overset(n)sumr`