Question

Find the sum of the cubes of the first $n$ natural numbers.

• A. $n^2(n+1{)}^2$
• B. $\left(n\right)(n+1)(2n+1)$
• C. $[\frac{n(n+1)}{2}{]}^2$
• D. None of these

Answer 1

The correct option is C

$[\frac{n(n+1)}{2}{]}^2$

We know that,

$r^4-(r-1{)}^4=4r^3-6r^2+4r-1$
Taking summation on both sides, then
${\sum }_{r=1}^n[r^4-(r-1{)}^4]=4{\sum }_{r=1}^n{r}^3-6{\sum }_{r=1}^n{r}^2+4{\sum }_{r=1}^{n}r-{\sum }_{r=1}^{n}1$

$n^4-0^4=4\sum n^3-6\sum n^2+4\sum r-n$

Substituting values of $\sum n^2$ and $\sum n$ in the above equation, we get

$n^4=4\sum n^3-\frac{6n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{6}-n$

$4\sum n^3=n^4+1(n+1)(2n+1)-2n(n+1)+n$

$=n[n^3+(n+1\left)\right(2n+1)-2(n+1)+1]$

$=n(n^3+2n^2+n)=n^2(n+1{)}^2$

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