If Sn denotes the sum of the cubes of the first n natural numbers and sn

Question:

If $S_{n}$ denotes the sum of the cubes of the first $n$ natural numbers and $s_{n}$

denotes the sum of the first n natural numbers then find the value of $\sum_{k=1}^{n} \frac{S_{k}}{S_{k}}$

Solution:

Given in the question, Sn denotes the sum of the cubes of the first n natural numbers.

sn denotes the sum of the first n natural numbers.

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 squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$

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

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

IV. Sum of a constant k, N times,

$\sum_{k=1}^{N} k=N k$

So,

$\mathrm{S}_{\mathrm{n}}$ denotes the sum of the cubes of the first $\mathrm{n}$ natural numbers. (Given data in the question).

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

sn denotes the sum of the first n natural numbers.

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

$\frac{S_{k}}{S_{k}}=\frac{\left(\frac{k(k+1)}{2}\right)^{2}}{\frac{k(k+1)}{2}}=\frac{k(k+1)}{2}$

To determine the given ratio in the question,

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

$=\frac{1}{2}\left[\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k\right]$

$=\frac{1}{2}\left[\left(\frac{n(n+1)(2 n+1)}{6}\right)+\left(\frac{n(n+1)}{2}\right)\right]$

$=\frac{n(n+1)}{4}\left[\left(\frac{(2 n+1)}{3}\right)+1\right]$

$=\frac{n(n+1)}{4}\left[\left(\frac{(2 n+4)}{3}\right)\right]$

$=\frac{n(n+1)(n+2)}{6}$

So, the value of $\sum_{k=1}^{n} \frac{s_{k}}{s_{k}}=\frac{n(n+1)(n+2)}{6}$