Sum of K positive integers

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Sum of K positive integers

by fruti_yum » Wed Aug 05, 2009 6:59 am
How to approach the attached question? In under 2 mins!!!


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by umaa » Wed Aug 05, 2009 7:18 am
The sum of the first k positive integers is k(k+1)/2

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

The sum of the m integers is, m(m+1)/2

Since, 0<n<m, The sum of the integers from n to m is, m(m+1)/2 - n(n+1)/2

IMO B.

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Re: Sum of K positive integers

by ankitns » Wed Aug 05, 2009 7:22 am
Since 0 < n < m

the sum of all integers from n to m (both inclusive) will be sum of first m integers - the sum if (n-1) integers.

If you plug in these into the formula...you would get

[m(m+1)/2] - [(n-1)n/2]

This will give us the answer C


Now to verify this lets say n = 3 and m = 5.

The answer should be 3 + 4 + 5 = 12

if we use the method above
A )sum of m numbers = 5 + 4 + 3 + 2 + 1 = 15.
B) sum of n-1 numbers = 2 + 1 = 3

Hence A - B = 15 - 3 = 12 (Bingo!!)
fruti_yum wrote:How to approach the attached question? In under 2 mins!!!


OA to be discussed!
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!