Question:
If $\sum_{k=1}^{n} k=45$, find the value of $\sum_{k=1}^{n} k^{3} .$
Solution:
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,
$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}=45^{2}=2025$