 Shirley Thompson

2021-12-26

The sum of the first n odd natural numbers is equal to ${n}^{2}$ poleglit3

Step 1
Let 1, 3, 5, 7, 9, n odd numbers be the first.
It is an Arithmatic Progression, whose first term $\left(a\right)=1$, and common difference (d) is 2.
Step 2
The total of these n terms $A\cdot P=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
first n odd numbers added together $=\frac{n}{2}\left[2×1+\left(n-1\right)×2\right]$
$=\frac{n}{2}\left[2+2n-2\right]$
$=\frac{n}{2}×n={n}^{2}$
Hence, the sum of first n odd natural numbers is equal to ${n}^{2}$. Hence, proved. nghodlokl

$S=1+3+5+\dots +\left(2n-5\right)+\left(2n-3\right)+\left(2n-1\right)$
is the sum of the first n odd numbers
We can also write
$S=\left(2n-1\right)+\left(2n-3\right)+\left(2n-5\right)+\dots +5+3+1$
$2S=\left[1+\left(2n-1\right]+\left[3+\left(2n-3\right)\right]+\left[5+\left(2n-5\right)\right]+\dots +\left[\left(2n-5\right)+5\right]+\left[\left(2n-3\right)+3\right]+\left[\left(2n-1\right)+1\right]$
$⇒2S=\left(2n\right)+\left(2n\right)+\left(2n\right)+\dots +\left(2n\right)+\left(2n\right)+\left(2n\right)$ n terms
$⇒2S=\left(2n\right)\left(n\right)$
$⇒2S=2{n}^{2}$
$⇒S={n}^{2}$
$⇒1+3+5+\dots +\left(2n-5\right)+\left(2n-3\right)+\left(2n-1\right)={n}^{2}$ user_27qwe

Proving by example
Sum of first 3 odd numbers is ${3}^{2}=9$
First 3 odd numbers are 1+3+5=9
Hence ,Sum of first n odd natural numbers is ${n}^{2}$

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