Find the sum of the first 50 positive integers.?

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Find the sum of the first 50 positive integers.?

2007-01-17 01:55:07 · 4 answers · asked by Bill B 1 in Science & Mathematics ➔ Mathematics

4 answers

50(50+1)/2 .

2007-01-17 06:47:09 · answer #1 · answered by tablecloth 1 · 0⤊ 0⤋

The sum of the first n positive integers is n(n+1)/2.

Why? This is equal to
(1+n) + (2+n-1) + (3+n-2) ... + (n/2 + n/2+1) if n is even

Each of the bracketed sums is equal to n+1, and there are n/2 such bracketed sums. So, (n+1) * n/2. The proof is a bit different for odd n, but the result is the same.

2007-01-17 01:59:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋

S=n/2(2a+(n-1)d)
S=50/2(2x1+49x1)
S=25(51)
S=1275

2007-01-17 02:08:42 · answer #3 · answered by Maths Rocks 4 · 0⤊ 0⤋

this is given by an equation
n(n+1)/2
replace n by 50
50*51/2 = 1275

2007-01-17 02:04:05 · answer #4 · answered by catarthur 6 · 0⤊ 0⤋