# proof by induction

• May 23rd 2012, 07:53 AM
Tweety
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,

• May 23rd 2012, 08:06 AM
Plato
Re: proof by induction
Quote:

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