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`

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 + 8 S_(1)) =

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

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

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

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

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

If S_1.S_2,S_3 are the sum of first n natural numbers,their squares and their cubes respectively,Show that : 9S_2^2=S_3(1+8S_1) .

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

If S_1, S_2, S_3 are the sum of first n natural numbers, their squares and their cubes, respectively , show that 9 S_(2)^(2) = S_(3) (1+ 8S_1)

If S_1, S_2, S_3 are the sum of first n natural numbers, their squares and their cubes, respectively , show that 9 S_(2)^(2) = S_(3) (1+ 8S_1)