Question

If $1+2+3+....+p=171$, then find the $1^3+2^3+3^3+....+p^3$

Answer 1

In the given series $1+2+3++....+p=171$, $S_p=171$ represents the sum of natural numbers upto $p$. From this we have to find sum of cubes of natural numbers upto $p$. Thus, we can write the formula instead of the series and that is:

Sum of natural numbers is $S_n=\frac{n(n+1)}{2}$, therefore, we have

$S_p=\frac{p(p+1)}{2}\Rightarrow 171=\frac{p(p+1)}{2}......\left(1\right)$

We know that the sum of cubes of $n$ natural numbers is $S_n={\left[\frac{n(n+1)}{2}\right]}^2$, thus consider the sum as follows:

$1^3+2^3+3^3+.......+p^3=S_p\Rightarrow 1^3+2^3+3^3+.......+p^3={\left[\frac{p(p+1)}{2}\right]}^2\Rightarrow 1^3+2^3+3^3+.......+p^3=\left(171{)}^2\right(from\left(1\right))\Rightarrow 1^3+2^3+3^3+.......+p^3=29241$

Hence $1^3+2^3+3^3+.......+p^3=29241$.