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

let `S_(n)` denote the sum of the cubes of the first n natureal number

1.	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
Answer» Correct Answer - `(1)/(6)n(n+1)(n+2)` `S_(k)={sum_(r=1)^(k)r^(3)}={(1)/(2)k(k+1)}^(2) and s_(k)={sum_(r=1)^(k)r}=(1)/(2)k(k+1).` `therefore (S_(k))/(s_(k))=({(1)/(2)k(k+1)}^(2))/((1)/(2)k(k+1))=(1)/(2)k(k+1)=(1)/(2)k^(2)+(1)/(2)k` `rArr sum_(k=1)^(n)(S_(k))/(s_(k))=(1)/(2)[sum_(k=1)^(n)k^(2)}+(1)/(2){sum_(k=1)^(n)k}` `=(1)/(2)xx(1)/(6)xxn(n+1)(2n+1)+(1)/(2)xx(1)/(2)n(n+1)` `=(1)/(12){n(n+1)(2n+1)+3n(n+1)}=(1)/(6)n(n+1)(n+2).`