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

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- Mar 21st 2011, 04:32 PMcompstudentMathematical 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 - Mar 21st 2011, 04:52 PMpickslides
So now use the assumption n=k to prove n=k+1

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

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

Using n=k for n=k+1 $\displaystyle \displaystlye 1+4+9+\dots +k^2 +(k+1)^2 = \frac{k(k+1)(2k+1)}{6}+(k+1)^2=\dots $ - Mar 21st 2011, 06:55 PMArchie Meade
However, to add to what pickslides has written,

you need to be aware of what you should end up with when you evaluate that sum!

You are attempting to show that

$\displaystyle \displaystyle\ 1+4+9+....+k^2+(k+1)^2=\frac{(k+1)(k+2)[2(k+1)+1]}{6}$

"if"

$\displaystyle \displaystyle\ 1+4+9+...+k^2=\frac{k(k+1)(2k+1)}{6}$