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Given that 1+2+.......+n=n(n+1)/2 and

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Given that 1+2+.......+n=n(n+1)/2 and

by Max@Math Revolution » Fri Jan 26, 2018 12:14 am
[GMAT math practice question]

Given that $$1+2+...+n=\frac{n\left(n+1\right)}{2}\ and\ 1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$ ,

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
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by Brent@GMATPrepNow » Fri Jan 26, 2018 7:44 am
Max@Math Revolution wrote:[GMAT math practice question]

Given that $$1+2+...+n=\frac{n\left(n+1\right)}{2}\ and\ 1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$ ,

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
Answer = (Sum of all integers from 1 to 100 inclusive) - (Sum of integers from 1 to 100 that are SQUARES of integers)

Note: 100 is the greatest SQUARE among the numbers from 1 to 100.
100 = 10²

So, answer = (100)(100 + 1)/2 - (10)(10 + 1)(2)(10 + 1)/6
= (100)(101)/2 - (10)(11)(21)/6
= 5050 - 385
= 4665
= B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Sun Jan 28, 2018 5:41 pm
=>
Since there are 10 squares of integers between 1 and 100, inclusive, we need to find
$$1+2+...+100-\left(1^2+2^2+...+10^2\right)=\frac{\left(100\cdot101\right)}{2}-\frac{\left(10\cdot11\cdot21\right)}{6}=5050-385=4665$$

Therefore, the answer is B.
Answer : B
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