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.


  • $\begingroup$ At least connect this to an even older post that is a target of several duplicate posts. $\endgroup$ Jan 11 '19 at 7:38

The induction step is just a matter of assuming that


and proving that


i.e., that


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


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

  • $\begingroup$ why did you add $(4k^3 -k)/3$ to the left side? $\endgroup$
    – Mac
    Oct 21 '13 at 0:14
  • $\begingroup$ @Mac: I didn’t. I used the induction hypothesis to replace $$1^2+3^2+\ldots+(2k-1)^2$$ by $$\frac{4k^3-k}3\;.$$ $\endgroup$ Oct 21 '13 at 0:18
  • $\begingroup$ I tried algebraically and I don't follow... I must be stupid $\endgroup$
    – Mac
    Oct 21 '13 at 0:31
  • $\begingroup$ @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. $\endgroup$ Oct 21 '13 at 0:39
  • $\begingroup$ thank you I got it after, btw i think you mean 3 instead of 1 on your expression for the left side. $\endgroup$
    – Mac
    Oct 21 '13 at 0:42

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 $(4(n+1)^3-(n+1))/3$.


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