In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n.$

role="math" localid="1648111669874" $1^2+2^2+3^2+...+n^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.n\left(n+1\right)\left(2n+1\right)$

The given statement is$1^2+2^2+3^2+...+n^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.n\left(n+1\right)\left(2n+1\right).$We need to prove that the given statement is true for all natural numbers by using mathematical induction.

Let's prove that the statement is true for $n=1.$

${\left(1\right)}^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.1\left(1+1\right)\left(2·1+1\right).$

$1=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\times 2\times 3.$

$=1.$

So, the statement is true for $n=1.$

Let's assume that the statement is true for $n=k.$

$1^2+2^2+3^2+...+k^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.k\left(k+1\right)\left(2k+1\right).$

Let's prove that the statement is true for $n=k+1.$

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

localid="1648112186639" $=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left(2k+1\right)+{\left(k+1\right)}^2.$

$=\left(k+1\right)\left[\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.k\left(2k+1\right)+\left(k+1\right)\right].$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left[k\left(2k+1\right)+6\left(k+1\right)\right].$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left(2k^2+k+6k+6\right).$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left(2k^2+7k+6\right).$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left(2k^2+4k+3k+6\right).$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right].$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left(2k+2\right)\left(2k+3\right).$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\left(k+1\right)\left[\left(k+1\right)+1\right]\left[2\left(k+1\right)+1\right].$

So, the statement is true for $n=k+1.$

Since the given statement is true for $n=1$and if the statement is true for some localid="1648112621601" $n=k$and it is also true for the next natural number$n=k+1.$

So by the principle of mathematical induction, the statement is true for all natural numbers.