How can we derive the formula for sum of odd numbers?

Question

We know that $\sum^n_{k=1}(2k-1)=n^2$. Can it be shown using AP?

Answer 1

Yes you just have to put the values.

$$\text{Sum = S}_n = \frac{n(2a+(n-1)d)}{2}$$ $$=\frac{n(2+(n-1)2)}{2}$$ $$=\frac{2n(n-1+1)}{2}={n^2}$$

Answer 2

$S_n = \frac{a_1 + a_n}{2} * n$
$a_1 = 1, a_n = 2n - 1 \Rightarrow S_n = \frac{2n}{2} * n = n^2$

Answer 3

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Answer 4

1st Method

$\displaystyle\sum_{k=1}^n\left(2k-1\right)=\displaystyle\sum_{k=1}^n2k-\displaystyle\sum_{k=1}^n1=2\displaystyle\sum_{k=1}^nk-n=\left(2 \cdot\dfrac{n(n+1)}{2}\right)-n=n^2$

2nd Method

$\displaystyle\sum_{k=1}^n\left(2k-1\right)=\displaystyle\sum_{k=1}^{2n-1}k-\displaystyle\sum_{k=1}^{n-1}2k=\dfrac{2n(2n-1)}{2}-n(n-1)=n(2n-1-n+1)=n^2$

3rd Method

$\displaystyle\sum_{k=1}^n\left(2k-1\right)=\displaystyle\sum_{k=0}^{n-1}\left((k+1)^2-k^2\right)=n^2$

Answer 5

Another method, similar to the one little Gauss probably used for the sum of all consecutive integers between $\;1\;$ and $\;n\;$ :

$$\begin{align*}\text{(ascending)}\;\;&S=&1&+&3&+&\ldots&+&(2n-3)&+&(2n-1)\\ \text{(descending)}\;\;&S=&(2n-1)&+&(2n-3)&+&\ldots&+&3&+&1\end{align*}$$

Now sum up summandwise both lines above:

$$2S=\overbrace{2n+2n+\ldots+2n}^{n\;\text{ times}}=2n^2$$

and we're done.

Answer 6

Use induction:

Base case: $$1=1^2$$

Step: $$1+3+...+2n-1=n^2 ⇒ 1+3+...+2n-1+2n+1=n^2+2n+1=(n+1)^2$$