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# The sum of integers from 1 to n (inclusive) is n(n+1)/2.

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Director
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The sum of integers from 1 to n (inclusive) is n(n+1)/2.  [#permalink]

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11 Dec 2005, 06:32
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The sum of integers from 1 to n (inclusive) is n(n+1)/2.
What is the sum from 50 to 100 (inclusive) ?
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Auge um Auge, Zahn um Zahn !

Director
Joined: 10 Oct 2005
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11 Dec 2005, 10:10
gamjatang wrote:
The sum of integers from 1 to n (inclusive) is n(n+1)/2.
What is the sum from 50 to 100 (inclusive) ?
It seem s like i saw this problem on this Forum already!
Sum from 1 to 50 is 50(50+1)/2=25(51)
Sum from 1 to 100 is 100(101)/2=50(101)
Sum from 50 to 100(liitle trap here I guess) is 50(101)-25(51)
50(101)-25(51)=25(2(101)-51)=25*151=3775
Should be 3775 if my math is correct
Director
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11 Dec 2005, 14:34
3825

Average of number from 50 to 100 is (50+100)/2=75

Number of in integer between 50 and 100 ( inclusive) = (50+100)/2 = 75

sum = average*no of integers = 75*51=3825
VP
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11 Dec 2005, 18:28
gamjatang wrote:
The sum of integers from 1 to n (inclusive) is n(n+1)/2.
What is the sum from 50 to 100 (inclusive) ?

Sum = (# of numbers)(average)
Sum = (100-50+1)(150/2) = 3,825
Senior Manager
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20 Dec 2005, 03:43
gamjatang wrote:
The sum of integers from 1 to n (inclusive) is n(n+1)/2.
What is the sum from 50 to 100 (inclusive) ?

50+51+..+100 = (1+...+100) - (1+2+...49)
= (100*101/2) - (49 * 50/2)
= 5050 - 1225
= 3825
Re: PS - Sum   [#permalink] 20 Dec 2005, 03:43
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