Question

Sum of cubes of the first $n$ natural numbers i.e., $1^3+2^3+3^3+4^3+\ldots \ldots+n^3$ is

• A. $\frac{[n(n+2)]^2}{4}$
• B. $\frac{[n(n+1)]^2}{4}$
• C. $\frac{[n(n-1)]^2}{4}$
• D. $\frac{[n(n+1)]^2}{2}$

Answer 1

Answer: (B)

Here, $S_n=1^3+2^3+3^3+\cdots+n^3$
$\because (k+1)^4-k^4=4 k^3+6 k^2+4 k+1$
Putting $k=1,2,3, \ldots, n$, we get
$ 2^4-1^4=4(1)^3+6(1)^2+4(1)+1 $
$3^4-2^4=4(2)^3+6(2)^2+4(2)+1$
$4^4-3^4=4(3)^3+6(3)^2+4(3)+1$
$.................$
$..............$
$.................$
$(n-1)^4-(n-2)^4 =4(n-2)^3+6(n-2)^2+4(n-2)+1 $
$n^4-(n-1)^4 =4(n-1)^3+6(n-1)^2+4(n-1)+1 $
$(n+1)^4-n^4 =4 n^3+6 n^2+4 n+1$
Adding both sides, we get
$(n+1)^4-1^4=$
$4\left(1^3+2^3+\ldots+n^3\right)+6\left(1^2+2^2+3^3\right. $
$ \left.+\ldots+n^2\right)+4(1+2+3+\ldots+n)+n$
$= 4 \displaystyle\sum_{k=1}^n k^3+6 \sum_{k=1}^n k^2+4 \sum_{k=1}^n k+n$
$\because \displaystyle\sum_{k=1}^n k=\frac{n(n+1)}{2}$ and $\displaystyle\sum_{k=1}^n k^2=\frac{n(n+1)(2 n+1)}{6}$
$\therefore 4 \displaystyle\sum_{k=1}^n k^3=n^4+4 n^3+6 n^2+4 n-6\left(\frac{n(n+1)(2 n+1)}{6}\right)$
$-4\left(\frac{n(n+1)}{2}\right)-n$
$\Rightarrow 4 S_n=n^4+4 n^3+6 n^2+4 n-n\left(2 n^2+3 n+1\right)$
$-2 n(n+1)-n$
$=n^4+2 n^3+n^2$
$=n^2\left(n^2+2 n+1\right)$
$=n^2(n+1)^2$
Hence,
$S_n=\frac{n^2(n+1)^2}{4}$
$S_n=\frac{[n(n+1)]^2}{4}$