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

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Intern
Joined: 09 Aug 2004
Posts: 11

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If the sum of k consecutive integers is represented by k(k+1 [#permalink]

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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

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Director
Joined: 18 Apr 2005
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Location: Canuckland

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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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Director
Joined: 18 Feb 2005
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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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SVP
Joined: 05 Apr 2005
Posts: 1705

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Re: PS: Interesting question [#permalink]

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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

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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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