Question

Solve : $1^3+2^3+3^3+...+n^3=\frac{1}{4}{n}^2(n+1{)}^2$

Answer 1

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

for n = 1

LHS = $1^3$

RHS $={\left(\frac{1(1+1)}{2}\right)}^2=1$

LHS = RHS

P(x) is true for n =1

P(k)

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

P(k+1)

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

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

$=\frac{k^2(k+1{)}^2+4(k+1{)}^3}{4}$

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

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

This $1^3+2^3{3}^3+...+k^3+(k+1{)}^3={\left(\frac{(k+1)(k+2)}{2}\right)}^2$ i.e;

P(k+1) is five whenever p(k) is the

$\therefore$ By the principle of mathematical on distron

P(n) is true for n where n is natural number