Question

For all $n\ge 1$, prove that $1^2+2^2+3^2+4^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$

Answer 1

Step $1:$ Assume given statement
Let the given statement be $P\left(n\right),$ i.e.
$P\left(n\right):1^2+2^2+3^2+4^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$

Step $2:$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1^2=\frac{1(1+1)(2\cdot 1+1)}{6}$
$\Rightarrow 1=\frac{1\cdot 2\cdot 3}{6}$
$\Rightarrow 1=1$
which is true for $n=1$.

Step $3:$ $P\left(n\right)$ for $n=k$.
Put $n=K$ in $P\left(n\right)$, and assume that $P\left(K\right)$ is true for some positive integer $K$ i.e.,
$P\left(K\right):1^2+2^2+3^2+4^2+\dots +K^2=\frac{K(K+1)(2K+1)}{6}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is also true whenever $P\left(K\right)$ is true.
Put $n=K+1$ in given statement.
$1^2+2^2+3^2+4^2+\dots +K^2+(K+1{)}^2$
$=\frac{K(K+1)(2K+1)}{6}+(K+1{)}^2$
$=\frac{K(K+1)(2K+1)+6(K+1{)}^2}{6}$
$=\frac{(K+1)(2K^2+K+6K+6)}{6}$
$=\frac{(K+1)(2K^2+7K+6)}{6}$
$=\frac{(K+1)(2K^2+4K+3K+6)}{6}$

$=\frac{(K+1)\left[2K\right(K+2)+3(K+2\left)\right]}{6}$
$=\frac{(K+1)(K+2)(2K+3)}{6}$
$=\frac{(K+1)(K+1+1)\left(2\right(K+1)+1)}{6}$
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 natural numbers $n$.