[GMAT math practice question]
1 + 2 + ... + k = k(k+1)/2. m and n are positive integers satisfying n < m. What is the value of n + (n+1) + ... + (m-1) + m?
A. (m+n)(m-n-1)/2
B. (m+n)(m-n+1)/2
C. (m-n)(m+n-1)/2
D. (m-n)(m+n+1)/2
E. (m+n)(m-n-1)/4
1 + 2 + … + k = k(k+1)/2. m and n are positive integers sa
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- Max@Math Revolution
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=>
n + (n+1) + ... + (m-1) + m
= 1 + 2 + ... + (n-1) + n + (n+1) + ... + (m-1) + m - (1 + 2 + ... + (n-1))
= m(m+1)/2 - (n-1)(n-1+1)/2
= m(m+1)/2 - n(n-1)/2
= (1/2)(m(m+1) - n(n-1))
= (1/2)(m^2 + m - n^2 + n)
= (1/2)(m^2 - n^2 + m + n )
= (1/2)( (m+n)(m-n) + (m+n) )
= (1/2)(m+n)(m-n+1)
Therefore, the answer is B.
Answer: B
n + (n+1) + ... + (m-1) + m
= 1 + 2 + ... + (n-1) + n + (n+1) + ... + (m-1) + m - (1 + 2 + ... + (n-1))
= m(m+1)/2 - (n-1)(n-1+1)/2
= m(m+1)/2 - n(n-1)/2
= (1/2)(m(m+1) - n(n-1))
= (1/2)(m^2 + m - n^2 + n)
= (1/2)(m^2 - n^2 + m + n )
= (1/2)( (m+n)(m-n) + (m+n) )
= (1/2)(m+n)(m-n+1)
Therefore, the answer is B.
Answer: B
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