# simple proof. LHS = RHS

• Oct 14th 2009, 03:23 PM
sirellwood
simple proof. LHS = RHS
If i have $\displaystyle \frac{1}{n}$$\displaystyle \sum_{k=1}^n k^2=\displaystyle \frac{1}{6}(n+1)(2n+1) where k=1,2,3....n how do i prove the LHS = RHS? • Oct 14th 2009, 03:55 PM redsoxfan325 Quote: Originally Posted by sirellwood If i have \displaystyle \frac{1}{n}$$\displaystyle \sum_{k=1}^n k^2$=$\displaystyle \frac{1}{6}$(n+1)(2n+1)

where k=1,2,3....n

how do i prove the LHS = RHS?

I claim that $\displaystyle \sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$.

How do I prove this? Induction.

1) It's clearly true for $\displaystyle n=1$.

2) Assume it's true for $\displaystyle n$.

3) Prove it's true for $\displaystyle n+1$:

$\displaystyle \sum_{k=1}^n k^2 +(n+1)^2=\frac{n(n+1)(2n+1)}{6}+(n+1)^2=$ ...

Spoiler:
$\displaystyle \frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{2n^3+3n^2+n}{ 6}+\frac{6(n^2+2n+1)}{6}=\frac{2n^3+9n^2+13n+6}{6}$ $\displaystyle =\frac{(n+1)(n+1)(2n+3)}{6}=\frac{(n+1)((n+1)+1)(2 (n+1)+1)}{6}=\sum_{k=1}^{n+1} k^2$

which is the desired result.

So $\displaystyle \sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$ and therefore $\displaystyle \frac{1}{n}\sum_{k=1}^n k^2=\frac{(n+1)(2n+1)}{6}$