### Home > PC > Chapter 8 > Lesson 8.2.6 > Problem8-149

8-149.

Use the mathematical induction to prove the following statement.

$1^2 + 2^2 + 3^2 + … + n^2 =\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$

$1^2 =\frac{1(1+1)((2)(1)+1)}{6}=\frac{(1)(2)(3)}{6}=\frac{6}{6}=1$

Assume $1^2 +...+ k^2=\frac{k(k+1)(2k+1)}{6}$ is true for some integer $k$.

$1^{2} + ... + k^{2} + \left(k + 1\right)^{2} =$
$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=$
$\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}=$
$\frac{(k+1)(k(2k+1)+6(k+1))}{6}=$
$\frac{(k+1)(2k^2+k+6k+6))}{6}=$
$\frac{(k+1)(k+2)(2k+3)}{6}$

So the induction step is true.

Since we know the statement is true for $n = 1$ and the induction step is true, we know that the statement is true for all positive integers $n$.