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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 [#permalink]

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[GMAT math practice question]

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?

\(A. 4050\)
\(B. 4665\)
\(C. 4775\)
\(D. 5000\)
\(E. 5050\)
[Reveal] Spoiler: OA

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Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink]

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New post 26 Jan 2018, 00:26
B

Sum of integer without the sum of the square of the numbers = sum of numbers up to 100 -sum of square of the numbers up to 10.

Hence the answer is 100(101)/2-10(11)(21)/6 =4665

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Given that 1+2+.......+n=n(n+1)/2 and [#permalink]

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New post 26 Jan 2018, 03:06
MathRevolution wrote:
[GMAT math practice question]

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?

\(A. 4050\)
\(B. 4665\)
\(C. 4775\)
\(D. 5000\)
\(E. 5050\)


The numbers between 1 and 100 which are squares are 1,4,9,16,25,36,49,64,81,100
The sum of these numbers is \(10 + 20 + 110 + 245 = 385\)
From the formula above, we can find the sum of all integers - \(\frac{100*101}{2} = 50*101 = 5050\)

Hence, the sum of integers between 1 and 100 that are not squares of integers are 5050 - 385 = 4665(Option B)
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Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink]

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New post 28 Jan 2018, 17:47
=>

Since there are 10 squares of integers between \(1\) and \(100,\) inclusive, we need to find

\(1+2+…+100 – (1^2+^22+…+10^2)
= \frac{(100*101)}{2} – \frac{(10*11*21)}{6} = 5050 – 385 = 4665\)

Therefore, the answer is B.
Answer : B
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Re: Given that 1+2+.......+n=n(n+1)/2 and   [#permalink] 28 Jan 2018, 17:47
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