# Formula for finding sequence of odd numbers starting from 0? [duplicate]

Let's say I had an odd number sequence: $$1+3+5+7+9+...+n=x$$ I was doing some problems and I coincidentally thought of: $x=(\frac{n+1}{2})^2$ I searched up the formula online to see if it actually works, and I didn't find any useful results. Wondering if it works?

The sum of any arithmetic sequence is $$\text{(number of terms)}\frac{\text{(first term)}+\text{(last term)}}{2}$$

For the sequence $1+3+5+7+9+...+n$:

\begin{align} \text{(number of terms)} &= \frac{n+1}{2} \\ \text{(first term)} &= 1 \\ \text{(last term)} &= n \end{align}

So $1+3+5+7+9+...+n = \left(\dfrac{n+1}{2}\right)\left(\dfrac{1+n}{2}\right) = \left(\dfrac{1+n}{2}\right)^2$

Proof by induction:

$1+3+5+...+n = \dfrac{(n+1)^2}{4}.$

0) $n=1:$ ok.

1) Assume formula valid for $n.$

2) Step: $n+2.$

(The next odd term is $n+2$).

$1+ 3+5 .....+ n + (n+2) =$

$\dfrac{(n+1)^2}{4} + (n+2) =$

$\dfrac{n^2+2n+1+4n+8}{4} =$

$\dfrac{((n+2)+1)^2}{4}.$

Your formula indeed works and it would be interesting to hear, how you came up with it..

This is how it's actually derived (Note that $n^{th}$ odd number is $2n-1$):

$$1+3+5+\cdots + (2n-1) = \sum_{i=1}^{n}\left[2i-1\right] = \sum_{i=1}^{n}2i - \sum_{i=1}^{n}1 = 2\sum_{i=1}^{n}i - \sum_{i=1}^{n}1$$

I assume you're familiar with the formula for $\sum_{i=1}^{n}i$. And If you sum $1$ $n$ times, you get $\sum_{i=1}^{n}1=n$:

$$2\sum_{i=1}^{n}i - \sum_{i=1}^{n}1=2(\frac{n(n+1)}{2})-n=n^2+n-n=n^2$$

We conclude that $1+3+5\cdots+(2n-1) = n^2$

• Actually, his formula does work. Call his $n$ $k$, then $k = 2n-1.$ So your $n^2 = \left(\frac{k+1}{2}\right)^2.$ – B. Goddard Sep 29 '17 at 5:17
• Oh, yeah. Of course. I guess I judged it too fast.. – Miksu Sep 29 '17 at 5:19
• You should use brackets for thr sum over $(2n-1)$ it might lead to confusions. – MrYouMath Sep 29 '17 at 5:40

$$\sum_{k=1}^n(2k-1)=\sum_{k=1}^n(k^2-(k-1)^2)=n^2-0^2=n^2.$$ From here $$1+3+5+...+n=1+3+...+2\cdot\frac{n+1}{2}-1=\left(\frac{n+1}{2}\right)^2.$$