Mathematical Induction

**compstudent** · March 21st 2011, 04:32 PM

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.

Help would be greatly appreciated

Thank you

**pickslides** · March 21st 2011, 04:52 PM

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

**Archie Meade** · March 21st 2011, 06:55 PM

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

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

"if"

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

Thread: Mathematical Induction

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