Question

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

Answer 1

$P\left(n\right)=1^3+2^3+3^3+\cdots +n^3={\left[\frac{n(n+1)}{2}\right]}^2$
For n = 1
$P\left(1\right)=1=1{\left[\frac{1(1+1)}{2}\right]}^2\Rightarrow 1=1$
$\therefore P\left(1\right)$ is true
Let P (n) be true for n = K
$\therefore P\left(k\right)=1^3+2^3+3^3+\cdots +K^3={\left[\frac{k(k+1)}{2}\right]}^{2}Forn=k+1P(k+1)=1^3+2^2+3^3+\cdots +k^3+(k+1{)}^3={\left[\frac{k(k+1)}{2}\right]}^2+(k+1{)}^3=(k+1{)}^2[\frac{k^2}{4}+k+1]=(k+1{)}^2\left[\frac{k^2+4k+4}{4}\right]=\frac{(k+1{)}^2(k+2{)}^2}{4}={\left[\frac{(k+1)(k+2)}{2}\right]}^2$
$\therefore$ P (k + 1) is true.
Thus P (k ) is true \Rightarrow (k +1) is true.
Hence by principle of mathematical induction,
P(n) is true for all $nϵN.$