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

Question

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?

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 06 '13 at 17:20
Related: http://math.stackexchange.com/questions/136237/direct-proof-that-1-3-5-cdots-2n-1-n-cdot-n/136239#136239 — Eric Naslund, Sep 06 '13 at 20:47

score 5 · Answer 1 · answered Sep 06 '13 at 19:18

5

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" :

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answered Sep 06 '13 at 19:18

Etienne

13,636

score 3 · Answer 2 · answered Sep 06 '13 at 17:18

3

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

answered Sep 06 '13 at 17:18

Umberto P.

52,165

Could you help me further. I am new with induction so I would appreciate your help. I just had induction with one variable. – 1mvdb Sep 06 '13 at 17:20
1

Can you prove that $1 + 3 + 5 + \cdots + (2n-1) = n^2$ using induction? – Umberto P. Sep 06 '13 at 17:23
3

Forget about $m$, it is just there to distract you. – André Nicolas Sep 06 '13 at 17:24
Thank you. This makes it a lot easier. – 1mvdb Sep 06 '13 at 17:26

score 2 · Answer 3 · answered Sep 06 '13 at 17:37

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

J. W. Perry · Accepted Answer · 2013-09-06T20:48:37.283

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

$$(1+3+...+(2k-1))=k^2.$$

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.

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

4 Answers4