An algebra problem by Shintaro Inaba

To solve this problem we must use the formula for sum of an arithmetic sequence, which is $\frac {(n(n + 1)} {2}$ where $n$ is the last term of the sequence. Plug in the values to get $\frac {(10000(10000 + 1)} {2} = \frac {(10000(10001)} {2} = \frac {100010000} {2} = 50005000$ . Thus the sum of all natural numbers from $1$ to $10000$ is equal to $50005000$ .

5000 *10001=50005000

Given $a = 1 , d = 1 , n = 10000 \text{ and last term} = 10000 \\ S = \dfrac{n}{2}(a + l) \\ = 5000 \times 10001 = \color{#20A900}{\boxed{50005000}}$