Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 2008:30 AM PDT
-09:30 AM PDT
Learn the five strategic moves that top MBA applicants make to unlock admission to M7 and Top 10 business schools. In this expert-led session, Sia Admissions — a leading coaching firm specializing in high-achieving candidates ... and more.
May 1610:00 AM PDT
-11:59 PM PDT
Save BIG with Magoosh! 50% off all study plans with code PREVAIL50 - Expires Monday 5/19- May 21
08:30 AM PDT
-09:30 AM PDT
How do you get into The Wharton School—one of the most prestigious MBA programs in the world?In this exclusive MBA Admissions Masterclass, former Wharton admissions directors Judith Silverman Hodara and Michel Belden break it down 
May 2211:30 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
May 2411:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn –1. Learn effective reading strategies 2.Tackle difficult RC & MSR with confidence 3.Excel in timed test environment
May 2408:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!- Jun 10
06:00 AM PDT
-05:30 PM PDT
Register for the GMAT Club Virtual MBA Spotlight fair, the biggest MBA fair of the year. You will have a chance to hear Admissions directors from almost every Top 20 program speak, network with peers, and conveniently attend morning or afternoon sessions.
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (01:55) correct
41%
(01:37)
wrong
based on 2605
sessions
History
Date
Time
Result
Not Attempted Yet
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. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
19 Jan 2012, 11:28
Kudos
Bookmarks
Baten80
The sum of the first k positive integers is equal to 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. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
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.
19 Jan 2012, 11:34
Kudos
Bookmarks
Baten80
The sum of the first k positive integers is equal to 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. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
Or try plug-in method: let m=4 and n=3 --> then m+n=7. Let see which option yields 7.
A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2} = 10-10=0\);
B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2} = 10-6=4\);
C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2} = 10-3=7\) --> OK;
D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2} = 6-10=-4\);
E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2} = 6-6=0\).
Answer: C.
