Find the sum!

the rule for consecutive integers is= $\frac{n (n+1)}{2}$

so, the summation is = $\frac{200 \times 201}{2}$ = $20100$

=200(200+1)/2

=40200/2

=20100

Relevant wiki: Arithmetic Progression Sum

It is an Arithmetic Progression with a common difference of $1$ . The desired sum is

$\color{#69047E}\huge{s=\dfrac{n}{2}(a_1+a_n)}\color{#3D99F6}=\dfrac{200}{2}(1+200)\color{#D61F06}=\boxed{20100}$

$\huge \color{#20A900}S \color{#D61F06}= \color{#3D99F6}\dfrac{n}{2} \color{#333333}\cdot \color{#BA33D6}(a + l) \color{#EC7300} \color{#E81990}= \color{#D61F06}\dfrac{200}{2} \color{#D61F06}\cdot \color{#BA33D6}(200 + 1) \color{#E81990}= \color{#69047E}100 \color{#D61F06}\times \color{#3D99F6}201 \color{#624F41}= \color{#D61F06}{\boxed{\color{#20A900}{20100}}}$

sum of this series
n(n+1)/2
in this case n is 200, so sum S will be
S = 200(200+1)/2
S = 20100

(A+l)/2 * n

There is a simple formula for finding the sum of first n natural numbers.

$\frac{n\times(n+1)}{2}$

Thus, using this formula for n = 200, we get the answer 20100