Question

If S 1 , S 2 , S 3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that

Answer 1

It is given that S 1 , S 2 , S 3 are the sum of first n natural numbers, their squares and their cubes respectively.

9 S 2 2 = S 3 ( 1+8 S 1 )(1)

Since S 1 is the sum of first n natural numbers,

S 1 = n( n+1 ) 2

Since S 2 is the sum of squares of first n natural numbers,

S 2 = n(n+1)( 2n+1 ) 6

Since S 3 is the sum of cubes of first n natural numbers,

S 3 = n 2 (n+1) 2 4

Substitute the values of S 1 , S 2 and S 3 to the R.H.S. of (1) we get,

S 3 ( 1+8 S 1 )= n 2 (n+1) 2 4 [ 1+ 8n(n+1) 2 ] = n 2 (n+1) 2 4 [ 1+4 n 2 +4n ] = [ n( n+1 )( 2n+1 ) ] 2 4

Consider L.H.S.,

9 S 2 2 = 9 ( n(n+1)( 2n+1 ) ) 2 6 2 = ( n(n+1)( 2n+1 ) ) 2 4

Hence, it is proved that 9 S 2 2 = S 3 ( 1+8 S 1 ).