Help with simple proof by mathematical induction

mcraze123
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Homework Statement



prove:
0^2 + 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6


Homework Equations





The Attempt at a Solution



I'm confused on how to prove this by induction. I'm not exactly sure what the goal of the rearrangement is after substituting (n+1). Any help is much appreciated!

base case: n = 0
0^2 = 0(0+1)(2*0+1)/6

induction step:
(0^2+1^2+2^2+...+n^2) + (n+1)^2 = (n+1)((n+1)+1)(2(n+1)+1)/6

n(n+1)(2n+1)/6 + (n+1)^2

(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...

Thanks!
 
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mcraze123 said:
(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...

Factor out 1/6 and then simplify what is left in the brackets.
 
mcraze123 said:
induction step:
(0^2+1^2+2^2+...+n^2) + (n+1)^2 = (n+1)((n+1)+1)(2(n+1)+1)/6

n(n+1)(2n+1)/6 + (n+1)^2

(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...
You're trying to show the first equation is true. To do this, you start with one side, as you have done, and manipulate it until it looks like the other side.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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