Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):1^3+2^3+3^3+\cdots +n^3={\left(\frac{n(n+1)}{2}\right)}^2$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1^3={\left(\frac{1(1+1)}{2}\right)}^2$
$\Rightarrow 1={\left(\frac{2}{2}\right)}^2$
$\Rightarrow 1=1$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for
some natural number $K$ i. e.,
$P\left(K\right):1^3+2^3+3^3+\cdots +K^3={\left(\frac{K(K+1)}{2}\right)}^2\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true
whenever $P\left(K\right)$ is true.
Now, we have
$1^3+2^3+3^3+\cdots +K^3+(K+1{)}^3$
$=(1^3+2^3+3^3+\cdots +K^3)+(K+1{)}^3$
$={\left(\frac{K(K+1)}{2}\right)}^2+(K+1{)}^3$ (Using $\left(1\right)$)
$=\left\{\frac{K^2(K+1{)}^2+4(K+1{)}^3}{4}\right\}$
$=\frac{(K+1{)}^2(K^2+4(K+1))}{4}$
$=\frac{(K+1{)}^2(K^2+4K+4)}{4}$
$=\frac{(K+1{)}^2(K+2{)}^2}{4}$
$={\left(\frac{(K+1)\left[\right(K+1)+1]}{2}\right)}^2$
Thus, $P(K+1)$ is true, whenever $P\left(K\right)$ is true.
Final Answer.
Hence, from the principle of Mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.