# 1+2+3+4........+100 = ?

2
by Tris11

2015-09-15T17:58:49+05:30
The sum of first 'n' natural numbers is [n(n+1)/2] .
Here , n = 100
⇒ 1+2+3+4+.......+100 = [100(100+1)/2]
⇒1+2+3+4+.....+100 = [100(101)/2]
⇒1+2+3+4+....+100 = 50(101)
⇒1+2+3+4+......+100 = 5050 .
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2015-09-15T18:07:28+05:30

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There is a formula for sum the firs t  N natural numbers from 1 onwards to N.
N = 100 here
Sum (1..N) = S_N = N (N + 1) / 2 = 100 * 101 /2 = 5050

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since the numbers to be added are having the same difference among them, we can also do the sum as:
[ the first number + last number ] /2 * number of terms
= [1 + 100 ] /2 * 100 = 5050

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