# simple proof. LHS = RHS

• Oct 14th 2009, 03:23 PM
sirellwood
simple proof. LHS = RHS
If i have $\frac{1}{n}$ $\sum_{k=1}^n k^2$= $\frac{1}{6}$(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 $\frac{1}{n}$ $\sum_{k=1}^n k^2$= $\frac{1}{6}$(n+1)(2n+1)

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

how do i prove the LHS = RHS?

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

How do I prove this? Induction.

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

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

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

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

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

which is the desired result.

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