Help with simple proof by mathematical induction

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SUMMARY

The discussion focuses on proving the formula for the sum of squares, specifically 0^2 + 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, using mathematical induction. The user presents their base case for n = 0 and outlines their induction step, which involves manipulating the expression (0^2 + 1^2 + 2^2 + ... + n^2) + (n+1)^2. They express confusion regarding the rearrangement and simplification needed to complete the proof. Key steps include factoring out 1/6 and simplifying the remaining expression.

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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.
 

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