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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
The sum of the first k positive integers is equal to k(k+1)/2. What is
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15 Nov 2021, 06:22
15 Nov 2021, 06:22
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The sum of the first k positive integers is equal to \(\frac{k(k+1)}{2}\). What is the sum of the integers from n to m, inclusive, where 0<n<m?
A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)
B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)
C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)
D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)
E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)
B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)
C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)
D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)
E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is
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24 Nov 2021, 10:42
24 Nov 2021, 10:42
Expert Reply
The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).
Answer: C.
Answer: C.
gmatclubot
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]
24 Nov 2021, 10:42
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