Question

If $1+2+3+....+n=k$ then $1^3+2^3+....+n^3$ is equal to

• A. $k^2$
• B. $k^3$
• C. $\frac{k(k+1)}{2}$
• D. $(k+1{)}^3$

Answer 1

The correct option is A $k^2$

In the given series $1+2+3++....+n=k$, $S_n=k$ represents the sum of natural numbers upto $n$. From this we have to find sum of cubes of natural numbers upto $n$. 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

$S_n=\frac{n(n+1)}{2}\Rightarrow k=\frac{n(n+1)}{2}......\left(1\right)$

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:

$1^3+2^3+3^3+.......+n^3=S_n\Rightarrow 1^3+2^3+3^3+.......+n^3={\left[\frac{n(n+1)}{2}\right]}^2\Rightarrow 1^3+2^3+3^3+.......+n^3=k^2\left(from\right(1\left)\right)$

Hence $1^3+2^3+3^3+.......+n^3=k^2$.