Question

Prove by induction:
$1^2+2^2+3^2+........+n^2=\frac{1}{6}n(n+1)(2n+1)$

Answer 1

The statement to be proved is:

$P\left(n\right):1^2+2^2+..+n^2=\frac{1}{6}n(n+1)(2n+1)$

Step 1: Verify that $P\left(n\right)$ is true for $n=1$

$P\left(1\right):\frac{1}{6}1(1+1)(2+1)$

$P\left(1\right):1=1$

Therefore, $P\left(1\right)$ is true.

Step 2: Assume that the statement is true for $n=k$

Let us assume that the below statement is true:

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

Step 3: Verify that the statement is true for $n=k+1$

We need to prove that:

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

$LHS=1^2+2^2+...+(k+1{)}^2$

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

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

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

$=\frac{k+1}{6}(2k^2+7k+6)$

$=\frac{k+1}{6}(2k^2+3k+4k+6)$

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

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

$=RHS$

Hence, $P\left(n\right)$ is true for all values of $n$ by principle of mathematical induction.