Prove that

It is given that, $\sum_{k=1}^{n} k=45$

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$

II. Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots . n^{3}$,

$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

From the above identities,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}=45$

We need to find,

$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}=45^{2}=2025$