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There is a simple formula that can be used to sum to a number. Where $n$ is the number you want to sum to.

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

$\frac { 100(101) }{ 2 }$

$\frac { 10100 }{ 2 } = 5050$

$1+2+3+\dotsm+(n-1)+n=\frac{n(n+1)}{2}$ Putting in $n=100$ ,we get: $1+2+3+\dotsm+99+100=\frac{100(100+1)}{2}=50\times101=\boxed{5050}$

How many numbers there from 1 to 100? The answer is 100. This means it's easy.

Example : take the numbers from 1 to 10. Here there are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. How many numbers there from 1 to 10? - 10. You can calculate them, but it'll take a long to calculate the numbers from 1 to 100. So you need something else.

1 + 10 = 11

2 + 9 = 11

3 + 8 = 11

4 + 7 = 11

5 + 6 = 11

5 * 11 = 11. Fine. But how can you do this for the numbers from 1 to 100? And how can you find how much times you need to multiply? Here's the answer. You can see above that the numbers end when they come consecutive. Like above 5 and 6. Maybe you can see the logic. You needed to multiply eleven 5 times and the last 11 if formed from 5 and 6. You need to multiply 11 with the smaller one - 5.

Now lets do it with the original problem.

1 + 100 = 101

...

50 + 51 = 101

50 * 51 = 5050

why is it so easy because: (1+100)x100:2=5050

n(n+1)/2, n=100, 100(101)/2=5050

= 101(50) = 5050

sum of natural number = (n*(n+1))/2