Question

Prove by Mathematical induction

$p\left(n\right)=\left\{1^3+2^3+3^3+....+n^3=\frac{n^2{\left(n+1\right)}^2}{4}\right\}$

Answer 1

To prove:-

$p\left(n\right)\cdot 1^3+2^3+3^3+.............+n^3=\frac{n^2(n+1{)}^2}{4}$

Proof by mathematical induction

When $n=1$

LHS :- $p\left(1\right)=1^3$

RHS:- $\frac{1(1+1{)}^2}{4}=\frac{1\times 2^2}{2^2}=1$

$\therefore p\left(1\right)$ is true.

Assume the result is true for $n=k$, that is

$1^3+2^3+3^3+..............+k^3=(1+2+3+..............+k{)}^2=(\frac{k(k+1)}{2}{)}^2$

Now for $n=k+1$, then

$1^3+2^3+3^3+.........+n^3$

$=1^3+2^3+3^3+.........+(k+1{)}^3$

$=(1+2+3+........+k{)}^2+(k+1{)}^3$

$=[\frac{k(k+1)}{2}{]}^2+(k+1{)}^3$

$=\frac{(k+1{)}^2}{4}(k^2+4k+4)$

$=\frac{(k+1{)}^2(k+2{)}^2}{4}$

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

$=(1+2+3+............+(k+1){)}^2$

Thus,

$1^3+2^3+3^3+4^3+..............+n^3=\frac{n^2(n+1{)}^2}{4}\forall n\in positiveintegers$