Pythagorean Odds

The sum first $n$ odd numbers is given by $n^2$ which in our case is $99^2=\boxed{9801}$

Let's say the sum is $\underbrace{1+3+5+\cdots\cdots+n}_{99~~\text{no. of terms}}$ where $n$ is ODD

It can be seen that the sum is in Arithmetic Progression (A.P) where each terms increments by $2$

Hence, we use the formula $S_n=\dfrac{n}{2}[2a+(n-1)d]$ where $n=\text{number of terms}, a=\text{first term}, d=\text{common difference}$

Derivation:

$S_n=\underbrace{[a]+[a+d]+[a+2d]+\cdots\cdots+[a+(n-1)d]}_{n~~\text{no. of terms}}~-~\text{(I)}$

Now we are going to write the sum in reverse order which gives us $S_n=\underbrace{[a+(n-1)d]+[a+(n-2)d]+\cdots\cdots+[a+d]+[a]}_{n~~\text{no. of terms}}~-~ \text{(II)}$

By $\text{(I)}+\text{(II)}$ we get:

$2S_n=\underbrace{[2a+(n-1)d]+[2a+(n-1)d]+\cdots\cdots+[2a+(n-1)d]}_{n~~\text{no. of terms}}$

Hence, $2S_n=n[2a+(n-1)d] \implies S_n=\dfrac{n[2a+(n-1)d]}{2}=\dfrac{n}{2}[2a+(n-1)d]$

Now back to the problem:

We saw that the sum is in A.P where each terms increments by $2$

Here $n=\text{number of terms}=99, a=\text{first term}=1, d=\text{common difference}=2$

Therefore, by putting the formua we get:

$\begin{aligned}S_n& =\dfrac{99}{2}[(2\times1)+(99-1)2] \\& =\dfrac{99}{2}[2+(98)2]\\& =\dfrac{99}{2}[2+196]\\& =\dfrac{99}{2}[198]\\& =\dfrac{99\times198}{2}=\boxed{9801}\end{aligned}$

Let, S=1+3+5+......(2n-5)+(2n-3)+(2n-1) S=(2n-1)+(2n-3)+(2n-5)......+5+3+1 S+S={(2n-1)+1}+{(2n-3)+3}+{(2n-5)+5}..... 2S=n(2n) S=n^2 Sum For first 99th odd numbers =99^2=99(100-1)=9900-99=9801 (Ans)

The sum of the first odd number is n², so it should be 99² = (90 + 9)² = 8100 + 1620 + 81 = 9801