Question

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

• A. $S_2^2$
• B. $9S_2$
• C. $9S_2^2$
• D. None

Answer 1

Answer: (C)

$9S_2^2$

$S_1=\frac{n(n+1)}{2}$
$1+8S_1=4n(n+1)+1=(2n+1{)}^2$
$S_2=\frac{n(n+1)(2n+1)}{6}$
$S_3={\left(\frac{n(n+1)}{2}\right)}^2$
Now,
$S_3\times (1+8S_1)$
$=\frac{(2n+1{)}^2(n^2\left)\right(n+1{)}^2}{4}$
$=9\times {\left(\frac{n(n+1)(2n+1)}{6}\right)}^2$
$=9(S_2{)}^2$
Hence, option C is correct.