# Math Help - proof by induction

1. ## proof by induction

prove by mathematical induction, that for n

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

assume that the summation formula is true for n=k

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

so must be true for n= k+1 ?

so do I put k+1 into the formula, and try and get it match the original? really stuck from this part,

2. ## Re: proof by induction

Originally Posted by Tweety
prove by mathematical induction, that for n

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

assume that the summation formula is true for n=k

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

so must be true for n= k+1 ?

so do I put k+1 into the formula, and try and get it match the original? really stuck from this part,
Note $\sum_{i=1}^{n+1} i^{2} =\sum_{i=1}^n i^{2} +(n+1)^2$