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pls help

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pls help

by tracyyahoo » Sun Oct 02, 2011 5:45 am
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. m(m+1)/2 - (n+1)(n+2)/2
B. m(m+1)/2 - n(n+1)/2
C. m(m+1)/2 - (n-1)n/2
D. (m-1)m/2 - (n+1)(n+2)/2
E. (m-1)m/2 - n(n+1)/2

why c not b,pls explain in details
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by sl750 » Sun Oct 02, 2011 6:23 am
1 2 3...n...m

Last term is m
Assume first term is n

m=n+k-1, where k is the number of terms between n and m inclusive
k=m-n+1

Sum = (m-n+1)/2(n+m) = (mn-n^2+n+m^2-mn+m)/2 = (m(m+1)-n(n-1))/2
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by knight247 » Sun Oct 02, 2011 6:30 am
This would be better understood with an example.
What is the sum of the 1st 5 digits i.e. 1 2 3 4 5?
5(5+1)/2=15

What is the sum of the 1st 7 digits i.e. 1 2 3 4 5 6 7
7(7+1)/2=28

Now, how would we find the sum of digits from 5 to 7 inclusive i.e. 5 6 7?
Sum of the first 7 digits-Sum of first 4 digits=7(7+1)/2 - 4(4+1)/2=28-10=18

The part u need to understand is that when u want the sum of digits from n to m inclusive, u need to eliminate any values before n. Hence it is C. Option B would be like finding the sum of digits 4 5 6 7 as per the example I cited.
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by GMATGuruNY » Sun Oct 02, 2011 8:09 am
tracyyahoo wrote:The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. m(m+1)/2 - (n+1)(n+2)/2
B. m(m+1)/2 - n(n+1)/2
C. m(m+1)/2 - (n-1)n/2
D. (m-1)m/2 - (n+1)(n+2)/2
E. (m-1)m/2 - n(n+1)/2

why c not b,pls explain in details
Let m=5 and n=3.
Sum of the integers 3 through 5 = 3+4+5 = 12. This is our target.
Now we plug m=5 and n=3 into the answers to see which yields our target of 12.

Answer choices A, B and C include m(m+1)/2:
m(m+1)/2 = 5(5+1)/2 = 15.
To yield our target of 12, we need to subtract 3.
Only answer choice C works:
(n-1)n/2 = (3-1)3/2 = 3.

Since m(m+1)/2 yields our target, the expression in D and E -- (m-1)m/2 -- will NOT yield our target.

The correct answer is C.
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