An algebra problem by Yahia El Haw

$\begin{aligned} S & = ({\color{#3D99F6}1}+100) + ({\color{#3D99F6}2}+99) + ({\color{#3D99F6}3}+98) + ({\color{#3D99F6}4}+97) + \cdots + ({\color{#3D99F6}50} +51) \\ & = \underbrace{101+101+101+101+ \cdots + 101}_{{\color{#3D99F6}50}\times 101} \\ & = \boxed{5050} \end{aligned}$

$S_{n} = \underbrace{ (1+n)+(2+n-1)+(3+n-2)+\cdots+\left(\dfrac{n}{2}+\dfrac{n}{2}+1 \right)}_{\frac{n}{2} \text{times}} \quad \quad \small\color{#3D99F6}{\text{For an even 'n'}} \\ S_{n} = \underbrace{ (1+n)+(1+n)+(1+n)+\cdots+(1+n)}_{\frac{n}{2} \text{times}} = \dfrac{n}{2}(n+1) \\ S_{100} = 50 \times 101 = 5050$

It is clearly seen that all the terms are $101$ . Since the number of terms is $50$ , the sum is $101(50)=5050$

S=(101*100)/2

This is equal to the $100$ th triangular number.

The formula for the $n$ th triangular number is $\frac{n(n+1)}{2}$ .

So the answer is $\frac{100*101}{2} = 5050$ .