Question

Sum of cubes of the first $n$ natural numbers i.e., $1^3+2^3+3^3+4^3+\dots \dots +n^3$ 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, $S_n=1^3+2^3+3^3+\cdots +n^3$
$\because (k+1{)}^4-k^4=4k^3+6k^2+4k+1$
Putting $k=1,2,3,\dots ,n$, we get
$2^4-1^4=4(1{)}^3+6(1{)}^2+4(1)+1$
$3^4-2^4=4(2{)}^3+6(2{)}^2+4(2)+1$
$4^4-3^4=4(3{)}^3+6(3{)}^2+4(3)+1$
$.................$
$..............$
$.................$
$(n-1{)}^4-(n-2{)}^4=4(n-2{)}^3+6(n-2{)}^2+4(n-2)+1$
$n^4-(n-1{)}^4=4(n-1{)}^3+6(n-1{)}^2+4(n-1)+1$
$(n+1{)}^4-n^4=4n^3+6n^2+4n+1$
Adding both sides, we get
$(n+1{)}^4-1^4=$
$4\left(1^3+2^3+\dots +n^3\right)+6(1^2+2^2+3^3$
$+\dots +n^2)+4(1+2+3+\dots +n)+n$
$=4\sum _{k=1}^n{k}^3+6\sum _{k=1}^n{k}^2+4\sum _{k=1}^{n}k+n$
$\because \sum _{k=1}^{n}k=\frac{n(n+1)}{2}$ and $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$
$\therefore 4\sum _{k=1}^n{k}^3=n^4+4n^3+6n^2+4n-6\left(\frac{n(n+1)(2n+1)}{6}\right)$
$-4\left(\frac{n(n+1)}{2}\right)-n$
$\Rightarrow 4S_n=n^4+4n^3+6n^2+4n-n\left(2n^2+3n+1\right)$
$-2n(n+1)-n$
$=n^4+2n^3+n^2$
$=n^2\left(n^2+2n+1\right)$
$=n^2(n+1{)}^2$
Hence,
$S_n=\frac{n^2(n+1{)}^2}{4}$
$S_n=\frac{[n(n+1){]}^2}{4}$