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Prove that $1^2 + 3^2 + 5^2+\cdots+(2n-1)^2 = (4n^3-n)/3$ for all $n \in \mathbb{N}$.

How can I solve this with induction? I've been working through a couple examples and for this one I can't relate the base case to the induction hypothesis.

I realize the base case is $n = 1$, which I check by putting $n = 1$ directly into $(4n^3-n)/3$ and $(2n-1)^2$, which proves the base case.

Then I tried to compose the last two terms of the sequence by: $\cdots + (2k-3)^2 + (2k-1)^2$. Am I approaching this correctly? How would I do this? A solution would be helpful as I've tried many other things such as expanding, making an inequality, etc.

Thanks!

Vedran Šego
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Mac
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2 Answers2

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The induction step is just a matter of assuming that

$$1^2+3^2+5^2+\ldots+(2k-1)^2=\frac{4k^3-k}3$$

and proving that

$$1^2+3^2+5^2+\ldots+(2k-1)^2+\big(2(k+1)-1\big)^2=\frac{4(k+1)^3-(k+1)}3\;,$$

i.e., that

$$1^2+3^2+5^2+\ldots+(2k-1)^2+(2k+1)^2=\frac{4(k+1)^3-(k+1)}3\;.\tag{1}$$

By the induction hypothesis the lefthand side of $(1)$ is equal to

$$\frac{4k^3-k}3+(2k+1)^2\;,\tag{2}$$

so you need only do the algebra necessary to show that $(2)$ is equal to the righthand side of $(1)$.

Brian M. Scott
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  • why did you add $(4k^3 -k)/3$ to the left side? – Mac Oct 21 '13 at 00:14
  • @Mac: I didn’t. I used the induction hypothesis to replace $$1^2+3^2+\ldots+(2k-1)^2$$ by $$\frac{4k^3-k}3;.$$ – Brian M. Scott Oct 21 '13 at 00:18
  • I tried algebraically and I don't follow... I must be stupid – Mac Oct 21 '13 at 00:31
  • @Mac: $(2)$ in my answer expands to $$\frac{4k^3-k}3+4k^2+4k+1=\frac{4k^3-k+12k^2+12k+3}3=\frac{4k^3+12k^2+11k+3}3$$ The righthand side of $(1)$ expands to $$\frac{4k^3+12k^2+12k+4-k-1}3=\frac{4k^3+12k^2+11k+3}3;,$$ which is the same. – Brian M. Scott Oct 21 '13 at 00:39
  • thank you I got it after, btw i think you mean 3 instead of 1 on your expression for the left side. – Mac Oct 21 '13 at 00:42
  • @Mac: You’re welcome. Yes, I did indeed mean $3$, and thanks to your comment, I caught it in time to edit my comment. – Brian M. Scott Oct 21 '13 at 00:44
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Assume the equation you displayed is correct for $n$. Add $(2(n+1)-1)^2$ to both sides. Show that the right-hand side equals $(4(n+1)^3-(n+1))/3$.

Stefan Smith
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