Starting with 1, at most how many consecuative positive integers can be added together BEFORE the sum exceeds 200? explain.
2007-09-25 21:24:25 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The general formula is 1+2+3+...+n=n(n+1)/2, so if you add up all the numbers from 1 to 10 you get 55. So what value of n gives you the largest sum not exceeding 200?
2007-09-25 21:30:51 · answer #1 · answered by zee_prime 6 · 0⤊ 0⤋
One could try to find the expression by hand. Yet the result of the sum is quite known. Those are the triangular number. They are given by these expression: n(n+1)/2.
You'll see that 19 is the answer.
You could do it still just by making the calculations mentally:P
2007-09-26 04:38:02 · answer #2 · answered by Mik 4 · 0⤊ 0⤋
n(n + 1)/2 ⤠200
Taking the equals,
n^2 + n - 400 = 0
n = (- 1 + â(1 + 1600))/2
approximating, n = [[- 1/2 + â400]] = [[- 1/2 + 20]]
n = 19
check:
19*20/2 = 190
20*21/2 = 210
2007-09-26 04:44:42 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
The formula: 1+2+3+...+n = n(n+1)/2
If we set:
n(n+1)/2 > 200
n(n+1) > 400
The smallest value for n is 20
2007-09-26 04:47:48 · answer #4 · answered by Christine P 5 · 1⤊ 0⤋
19..
use calculator!
2007-09-26 04:28:37 · answer #5 · answered by ^^' 2 · 1⤊ 0⤋