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Here is an induction proof that I have written for my homework and I want to know if I am understanding this correctly:

Prove that for:

$ \sum\limits_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

My proof:

Check the base-case, let mine be n=1 (or do I have to choose 0..? The lowest one)

$ (1^2) = \frac{1^2(1^2+1)(2 \cdot 1^2 +1)}{6} = \frac{6}{6} = 1 $

Base case proved.

Induction step. Assume that

$ (1^2+2^2=3^2...k^2) = \frac{k(k+1)(2k+1)}{6} $

works for all k.

Add another sum, $k+1$, then:

$(1^2+2^2=3^2...+k^2+(k+1)^2) = \frac{k(k+1)(2k+1)}{6}+(k+1) $

Substitude for sum up until $(k+1)^2$

$ \frac{k(k+1)(2k+1)}{6}+(k+1)^2 = \frac{k(k+1)(2k+1)}{6}+(k+1) $

Re-arrange and tidy up:

$ \frac{(k^2+k)(2k+1)+6k+6}{6} = \frac{(k^2+k)(2k+1)(6k+6}{6} $

Therefore we have proven that it works for $i=1$ and as it works for any $k+1$ it must work for the rest.

Is this correctly executed and understood?

EDIT: As my method seems to be wrong, let me have another try:

Now with:

$ \sum\limits_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

$(1^2+2^2=3^2...+k^2+(k+1)^2) = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} $

Re arrange and fiddle:

$2n^3+9n^2+13n+6 = 2n^3+9n^2+13n+6$

Is this correct?

Paze
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    $1^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$ must be shown to equalize $\frac{(k+1)(k+2)(2k+3)}{6}$ You work incorrectly with $1^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)$ – drhab Jan 23 '14 at 18:45
  • Additionally, you should not assume that it works for all $k.$. Assuming that it works for some $k$ is what is needed in the induction step. – Cameron Buie Jan 23 '14 at 18:51
  • What do you mean? Do I need to make 3 equalities...??? And why do I need to show that it equals that other equality (where I assume you put k+1 instead of n)? – Paze Jan 23 '14 at 18:53
  • Is my 2nd try in the edited post correct? – Paze Jan 23 '14 at 19:01

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If it is to be shown by induction that $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$ then $f\left(0\right)=g\left(0\right)$ must be proved (the base case) and secondly it must be shown that: $$\sum_{i=0}^{n+1}f\left(i\right)=g\left(n+1\right)$$ on base of: $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$ That amounts in showing that: $$g(n)+f\left(n+1\right)=g\left(n+1\right)$$ Note that here: $$g(n)+f\left(n+1\right)=\sum_{i=0}^{n+1}f\left(i\right)$$

In your case that comes to proving that: $$\frac{1}{6}n\left(n+1\right)\left(2n+1\right)+\left(n+1\right)^{2}=\frac{1}{6}\left(n+1\right)\left(n+2\right)\left(2n+3\right)$$

drhab
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  • I edited my post, is it now correct (the 2nd part) ? – Paze Jan 23 '14 at 18:58
  • Yes, only I should change the $k$ into a $n$ – drhab Jan 23 '14 at 19:01
  • I am lacking answers in my other thread (which inspired this one), could you have a look at it? http://math.stackexchange.com/questions/649025/bernoullis-inequality-by-induction#649025 – Paze Jan 23 '14 at 19:02