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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
18 Jun 2018, 03:43
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:32) correct
37%
(02:21)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
Given that \(1+2+...+n=\frac{n(n+1)}{2}\) and \(1^2+2^2+…+ n^2=\frac{n(n+1)(2n+1)}{6}\), what is the sum of the integers between 1 and 100 (inclusive) that are not squares of integers?
A. 4050
B. 4665
C. 4775
D. 5000
E. 5050
A. 4050
B. 4665
C. 4775
D. 5000
E. 5050
18 Jun 2018, 03:43
Kudos
Bookmarks
Official Solution:
Given that \(1+2+...+n=\frac{n(n+1)}{2}\) and \(1^2+2^2+…+ n^2=\frac{n(n+1)(2n+1)}{6}\), what is the sum of the integers between 1 and 100 (inclusive) that are not squares of integers?
A. 4050
B. 4665
C. 4775
D. 5000
E. 5050
Since there are 10 squares of integers between 1 and 100, inclusive, we need to find \(1+2+…+100 – (1^2+2^2+…+10^2)= \frac{(100*101)}{2} – \frac{(10*11*21)}{6} = 5050 – 385 = 4665\)
Answer: B
Given that \(1+2+...+n=\frac{n(n+1)}{2}\) and \(1^2+2^2+…+ n^2=\frac{n(n+1)(2n+1)}{6}\), what is the sum of the integers between 1 and 100 (inclusive) that are not squares of integers?
A. 4050
B. 4665
C. 4775
D. 5000
E. 5050
Since there are 10 squares of integers between 1 and 100, inclusive, we need to find \(1+2+…+100 – (1^2+2^2+…+10^2)= \frac{(100*101)}{2} – \frac{(10*11*21)}{6} = 5050 – 385 = 4665\)
Answer: B
05 May 2022, 05:47
Kudos
Bookmarks
just work on the formula and forget about its derivation. Its neither ap nor gp. We know the summation formulas for both but not for this one (out of scope).By dwelling on this, we dont dwell on the concept behind the solution, rather we go tangential in our attempt to solve the problem.
