40th Problem

$\large \displaystyle \text{Sum of n off natural numbers } 1,3,5,...... (2n-1).\\ \large \displaystyle \implies \sum_{i=1}^n (2n-1) = 2 \times \frac{n(n+1)}{2} - n = n^2 + n - n = n^2.\\ \large \displaystyle \text{Sum of odd numbers is } n^2 \\ \large \displaystyle \text{Sum of first 50 Odd numbers } = n^2 = 50^2 = \color{#D61F06}{\boxed{2500}}.$

$50th term = 1 + (50-1)2$

$= 1 + 98 = 99$

$S = n/2 * (a+l)$

$S = 25 * (1 + 99)$

$S = 25 * 100 = 2500$

Using the basic concepts of an Arithmetic Progression. https://brilliant.org/wiki/arithmetic-progressions/

This is an A.P. where the first term = 1, last term = 99, and there are 50 terms. S =(n/2)(a + l) = 25*100 = 2500

We know that sum of the first 'n' odd natural numbers is n^2. So , using that we get the sum of first 50 odd natural numbers = 50^2 = 2500.