# Prove the following by using the principle of mathematical induction for all $n\in N$:Q : 2        $1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2$

G Gautam harsolia

Let the given statement be p(n) i.e.
$p(n):1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2$
For n = 1  we have
$p(1):1=\left (\frac{1(1+1)}{2} \right )^2= \left ( \frac{1(2)}{2} \right )^2=(1)^2=1$    ,   which is true

For  n = k  we have
$p(k):1^3+2^3+3^3+...+k^3=\left (\frac{k(k+1)}{2} \right )^2 \ \ \ \ \ \ \ \ \ \ - (i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):1^3+2^3+3^3+...+(k+1)^3=1^3+2^3+3^3+...+k^3+(k+1)^3$
$=(1^3+2^3+3^3+...+k^3)+(k+1)^3$
$=\left ( \frac{k(k+1)}{2} \right )^2+(k+1)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k^2(k+1)^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^2+2ab+b^2=(a+b)^2)$
$=\left ( \frac{(k+1)(k+2)}{2} \right )^2$

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

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