Shirley Thompson

2021-12-26

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

poleglit3

Beginner2021-12-27Added 32 answers

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 $A\cdot P=\frac{n}{2}[2a+(n-1)d]$

first n odd numbers added together $=\frac{n}{2}[2\times 1+(n-1)\times 2]$

$=\frac{n}{2}[2+2n-2]$

$=\frac{n}{2}\times n={n}^{2}$

Hence, the sum of first n odd natural numbers is equal to $n}^{2$. Hence, proved.

nghodlokl

Beginner2021-12-28Added 33 answers

is the sum of the first n odd numbers

We can also write

Adding both results in sequence

user_27qwe

Skilled2022-01-05Added 375 answers

Proving by example

Sum of first 3 odd numbers is

First 3 odd numbers are 1+3+5=9

Hence ,Sum of first n odd natural numbers is

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