Question

If $s_1=\sum n,S_2=\sum n^2,S_3=\sum n^3$ then $9S_2^2=S_3(1+8S_1)$

• A. True
• B. False

Answer 1

The correct option is A True

Here, $S_1$ = Sum of first n natural numbers = $\frac{n(n+1)}{2}$ ---------(1)

$S_2$ = Sum of the squares of first n natural numbers = $\frac{n(n+1)(2n+1)}{6}$ ---------(2)

$S_3$ = Sum of the squares of first n natural numbers = ${\left[\frac{n(n+1)}{2}\right]}^2$ ---------(3)

Now, LHS = $9S_2^2$

$=9{\left[\frac{n(n+1)(2n+1)}{6}\right]}^2=9\frac{n^2{(n+1)}^2{(2n+1)}^2}{36}=9\times \frac{n^2{(n+1)}^2}{4}\times \frac{{(2n+1)}^2}{9}={\left[\frac{n(n+1)}{2}\right]}^2[9\times \frac{{(2n+1)}^2}{9}]={\left[\frac{n(n+1)}{2}\right]}^2[4n^2+4n+1]={\left[\frac{n(n+1)}{2}\right]}^2[1+4n(n+1)]={\left[\frac{n(n+1)}{2}\right]}^2[1+8n\frac{(n+1)}{2}]$

From (1) and (2), we get

$=S_3(1+8S_2)$ = RHS