Let `S_n` denote the sum of the cubes of the first `n` natural numbers and `s_n` denote the sum of the first `n` natural numbers. Then `sum_(r=1)^n S_r/s_r` is equal to

Let S_(n) denote the sum of the cubes of the first n natural numbers and s_(n) denote the sum of the first n natural numbers. Then sum_(r=1)^(n)(S_(r))/( S_(r )) equals :

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

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

Let S_(n) denote the sum of the cubes of the first n natural numbers and s_(n) denote the sum of first n natural numbers then (S_(n))/(s_(n)) is equal to (i) (n(n+1)(n+2))/6 (ii) (n( n+1))/2 (iii) (n^(2)+3n+2)/2 (iv) (2n+1)/3

Let S_(n) denote the sum of the cubes of the first n natural numbers and S'_(n) denote the sum of the first n natural numbers, then underset(r=1)overset(n)Sigma ((S_(r))/(S'_(r))) equals to

The sum of cubes of first n natural numbers is ……..of the first n natural numbers.

If S_(n) denotes the sum of the products of the products of the first n natureal number taken two at a time

If S_n denotes the sum of the first n terms of an AP and if S_2n=3S_n then S_3n:S_n is equal to