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 square

1.	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)=`
Answer» We have `S_(1)=(1+2+3+...+n)rArr S_(1)=(1)/(2)n(n+1),` `S_(2)=(1^(2)+2^(2)+3^(2)+ ...+n^(2))rArr S_(2)=(1)/(6)n(n+1)(2n+1),` `and S_(3)=(1^(2)+2^(3)+3^(3)+ ...+n^(3)) rArr S_(3)=(1)/(4)n^(2)(n+1)^(2)`. `therefore 9S_(2)^(2)=9xx(1)/(36)n^(2)(n+1)^(2)(2n+1)^(2)=(1)/(4)n^(2)(n+1)^(2)(2n+1)^(2).` And , `S_(3)(1+8S_(1))=(1)/(4)n^(2)(n+1)^(2){1+8*(1)/(2)n(n+1)}` ` =(1)/(4)n^(2)(n+1)^(2)(4n^(2)+4n+1)=(1)/(4)n^(2)(n+1)^(2)(2n+1)^(2).` Hence, `9S_(2)^(2)=S_(3)(1+8S_(1)).`