If S1 , S2 , S3 are the sums of first n natural numbers, their squares and their cubes respectively, then S3 (1 + 8S1 ) = 

Correct option is (A) \(9S_2\,^2\)

\(S_1=1+2+3+........+n=\frac{n(n+1)}2\)          ________________(1)

\(S_2=1^2+2^2+3^2+........+n^2=\frac{n(n+1)(2n+1)}6\)   ________________(2)

\(S_3=1^3+2^3+3^3+........+n^3=\left(\frac{n(n+1)}2\right)^2\)   ________________(3)

Now, \(S_3(1+8S_1)\) \(=\left(\frac{n(n+1)}2\right)^2\left(1+\frac{8n(n+1)}2\right)\)  (From (1) & (3))

\(=\left(\frac{n(n+1)}2\right)^2\left(1+4n^2+4n\right)\)

\(=\frac{(n(n+1))^2}4\,\left(2n+1\right)^2\)

\(=\frac{(n(n+1)(2n+1))^2}4\)

\(=\frac{(6S_2)^2}4\)                   (From (2))

\(=\frac{36S_2\,^2}4\) \(=9S_2\,^2\)

Correct option is A) 9S₂²