# Mathematical Induction

• March 21st 2011, 04:32 PM
compstudent
Mathematical Induction
I am having problems solving the second step. I have proven that n=1 in the first step.

I have to prove each given statement through mathematical induction.

http://i53.tinypic.com/34phphc.gif

Help would be greatly appreciated (Nod) Thank you
• March 21st 2011, 04:52 PM
pickslides
So now use the assumption n=k to prove n=k+1

n=k $\displaystlye 1+4+9+\dots +k^2 = \frac{k(k+1)(2k+1)}{6}$

n=k+1 $\displaystlye 1+4+9+\dots +k^2 +(k+1)^2$

Using n=k for n=k+1 $\displaystlye 1+4+9+\dots +k^2 +(k+1)^2 = \frac{k(k+1)(2k+1)}{6}+(k+1)^2=\dots$
• March 21st 2011, 06:55 PM
$\displaystyle\ 1+4+9+....+k^2+(k+1)^2=\frac{(k+1)(k+2)[2(k+1)+1]}{6}$
$\displaystyle\ 1+4+9+...+k^2=\frac{k(k+1)(2k+1)}{6}$