Property 3 The sum of the cubes of first n natural numbers is equal to the square of their sum. That is `1^3 + 2^3 + 3^3 +.........+ n^3 = (1+2+3+........+n)^2`

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 to

If S_(1),S_(2),S_(3) are the sums of first n natural numbers,their squares and their cubes respectively then S_(3)(1+8S_(1))=

If S_1, S_2, S_3 are the sums of first n natural numbers, their squares and cubes respectively, show that 9S_2^ 2=S_3(1+8S_1)dot

The sum of squares of first n natural numbers is given by 1/6 n (n+1)(2n+1) or 1/6 (2n^3 + 3n^2 + n) . Find the sum of squares of the first 10 natural numbers.