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If the sum of k consecutive integers is represented by k(k+1

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Intern
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If the sum of k consecutive integers is represented by k(k+1 [#permalink] New post 26 May 2005, 11:39
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If the sum of k consecutive integers is represented by k(k+1)/2 what is the sum of integers between n and m where 0<n<m
Director
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 [#permalink] New post 26 May 2005, 13:29
say f(k) = k(k+1)/2

then Sum from n to m = f(m) - f(n-1)
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 [#permalink] New post 26 May 2005, 14:42
first term = m

last term =n


first find the number of terms N (assumption is they are all consequtive)

n+(N-1)*1 = m ==> num of terms = m-n+1

Now find the sum ofthe series between n and m

Sum = N/2{2*first term + (N-1) difference between 2 terms)

= (m-n+1)/2{2*n +(m-n)*1}

==> (m-n+1)(m+n)/2
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Re: PS: Interesting question [#permalink] New post 26 May 2005, 18:03
iris wrote:
If the sum of k consecutive integers is represented by k(k+1)/2 what is the sum of integers between n and m where 0<n<m


the question seems incomplete. it should be : If the sum of consecutive integers [b]from 1 to k is represented by k(k+1)/2 what is the sum of integers between n and m where 0<n<m.

if so the sum = (m-n-1)(m+n)/2
Re: PS: Interesting question   [#permalink] 26 May 2005, 18:03
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If the sum of k consecutive integers is represented by k(k+1

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