# Verify the result of the sumVerify the result of the sum 1/100+2/100+...+99/100=99/2

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1/100 + 2/100 + 3/100 + ....+ 99/100

Since the denominators are the same, then we will add the numerators.

==> (1+2+3+...+99)/100

Now we notice that the numerator is a sum of a series

Then we will use the formula (n+1)*n/2

n= 99

==> (1+2+...+99) = (1+99)*99/2 = 50*99

Now we will substitute into the problem.

(1+2+...+99)/100 = 99*50/100 = 99/2............q.e.d

We have to verify the result of the sum 1/100+2/100+...+99/100 = 99/2.

Now we see that all the terms in 1/100+2/100+...+99/100 have the same denominator. So we can just add the numerators.

(1+ 2 + 3..99)/100

The sum of the first n numbers is n*(n+1)/2

=> (99*100/2)/100

=> 99/2

So we prove that 1/100+2/100+...+99/100 = 99/2.

First, we notice that the fractions from the left side have the same denominator, so we can add the numerators:

(1 + 2 + 3 + ... + 99)/100 = 99/2

We'll apply the formula of adding the first n terms:

1 + 2 + 3 + ... + n = (n+1)*n/2

1 + 2 + 3 + ... + 99 = (99+1)*99/2

1 + 2 + 3 + ... + 99 = 100*99/2

1 + 2 + 3 + ... + 99 = 50*99

Now, we'll substitute the sum from numerator, by the result:

50*99/100 = 99/2

We'll simplify by 50:

99/2 = 99/2

**The identity is true.**