Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this:

Prove for all $m, n \in \mathbb N$: $$[1 + 3 + \cdots + (2n - 1)]^m = n^{2m}$$

For $n = 1: 1 = 1^2$, hence P(1) is true.

Let $N \in \mathbb N$ be given and assume: $$[1 + 3 + \cdots + (2N - 1)]^m = N^{2m}$$

For $n = N + 1$: $$[1+3+\cdots+(2N-1)+(2(N+1)-1)]^m = (N+1)^{2m}$$

But I don't know what to do know or did I already made a mistake?

share|cite|improve this question
You can find an explicit formula for the sum inside by looking at $\sum_{i=1}^n i$ as the sum of the evens and the odds. That should help – Prahlad Vaidyanathan Sep 6 '13 at 17:20
up vote 1 down vote accepted

Umberto has the key, and I feel like typing. I hope I am not spoiling someone's homework.

$\textbf{Theorem.} \, \forall \, m,n \in \mathbb{N}, (1+3+...(2n-1))^m=n^{2m}.$

$\textbf{Proof.}$ We begin by proving that $(1+3+...+(2n-1))=n^{2},$ by induction on $n$, and then conclude that $(1+3+...+(2n-1))^m=n^{2m}$ by properties of exponents.

For for the base case let $n=1$. Then $$(2 \cdot 1-1)=1=1^2.$$

Assume the series holds for $n=k$. Then


For the inductive step, we let $n=k+1$ to show that $$(1+3+...+(2k-1)+(2k+1))=(k+1)^2.$$ On the left side of the equality we have, $$ \begin{align*} (1+3+...+(2k-1)+(2k+1)) &= (1+3+...+(2k-1)) + (2k+1) \\ &= k^2 + (2k+1) \\ &= k^2+2k+1 \\ &= (k+1)^2. \end{align*} $$

Thus we have shown that $$(1+3+...(2n-1))=n^{2}.$$

Now by exponentiation on both sides by $m$,

$$\qquad\qquad (1+3+...(2n-1))=n^{2} \Rightarrow (1+3+...(2n-1))^m=n^{2m}. \qquad\qquad \blacksquare$$

Please feel free to refine this, I no longer feel like typing.

share|cite|improve this answer

Apart from the standard proof by induction for the sum of the first odd numbers $1+3+\cdots +(2n-1)$, I'm pretty sure that you are all aware of the beautiful "picture proof" :

enter image description here

share|cite|improve this answer

Prove it for $m=1$ using induction on $n$. Then exponentiate.

share|cite|improve this answer
Could you help me further. I am new with induction so I would appreciate your help. I just had induction with one variable. – anonymous0543212 Sep 6 '13 at 17:20
Can you prove that $1 + 3 + 5 + \cdots + (2n-1) = n^2$ using induction? – Umberto P. Sep 6 '13 at 17:23
Forget about $m$, it is just there to distract you. – André Nicolas Sep 6 '13 at 17:24
Thank you. This makes it a lot easier. – anonymous0543212 Sep 6 '13 at 17:26

Note that $$1+2+\cdots+n=\frac{n(n+1)}{2}$$

Thus, $$1+3+\cdots+(2n-1) = (1+2+\cdots+(2n))-(2+4+\cdots+(2n))$$ $$=(1+2+\cdots+(2n))-2(1+2+\cdots+n)$$ $$=\frac{(2n)(2n+1)}{2}-2\cdot\frac{n(n+1)}{2}$$ $$=n^2$$

Therefore, $$\{1+3+\cdots+(2n-1)\}^m=(n^2)^m=n^{2m}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.