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: