How do you find the sum of the first 5000 natural numbers?

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Answer 1 · 2016-07-02T01:35:44

The sum of the first $n$ $n$ positive integers, $n$ $n$ being a natural number, is given by $S_{n} = \frac{n (n + 1)}{2}$ $S_n = (n(n + 1))/2$ .

$S_{5000} = \frac{5000 (5001)}{2}$ $S_5000 = (5000(5001))/2$

$S_{5000} = 12 502 500$ $S_5000 = "12 502 500"$

Hopefully this helps!

Answer 2 · 2016-07-02T02:16:59

12502500

Without using formula we can proceed as follows.

Let the sum be S then

$S= 1+2+3+4+5+ ---4998+4999+5000....(1)$

Again writing in reverse order

$S=5000+4999+4998+ ----+3+2+1......(2)$

Adding (1) and(2) we get

$2S=(5000+1)+(4999+2)"+"--+(4999+2)+(5000+1)$

$S=(5000xx5001)/2=12502500$