Sum of Natural Numbers

Note that $a_{k} + a_{n - k}$ is constant:

$1 + 200 = 2 + 199 = \ ... \ = 100 + 101 = 201$

Since after $100 + 101$ we will just have the same pair of numbers as the first half of the set, we do not count them in our calculation of the sum. Thus, there are $\frac{200}{2} = 100$ of these iterations of $a_{k} + a_{n - k}$ .

So our sum is

$S = 201 \times 100 = 20100$

This process of summing up consecutive numbers separated the same constant $($ i.e. a set of even numbers, odd numbers, multiples of $3$ , the first $n$ natural numbers, etc. $)$ is where the formula $\displaystyle \frac{n(n + 1)}{2}$ comes from.

1+200=201

2+199=201

3+198=201

................................

..................................

100+101=201

so, we get 100 pairs of 201 and the result is =(201 x 100)=20100

Using the formula given, the sum of the first 200 positive integers $1 + 2 + 3 +···+ 198 + 199 + 200$ = $\frac{200(200+1)}{2}$ $= 100(201) = 20100$

$\frac{200^2*200}{2}=20100$