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The sum of the first n positive perfect squares, where n is [#permalink]
 17 May 2012, 13:43
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The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares? A. 1,010 B. 1,164 C. 1,240 D. 1,316 E. 1,476
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Re: What is the sum of the first 15 positive perfect squares? [#permalink]
17 May 2012, 14:50
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The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?A. 1,010 B. 1,164 C. 1,240 D. 1,316 E. 1,476 Given that \frac{n^3}{3}+c*n^2+\frac{n}{6} gives the sum of the first n positive perfect squares. Now, for n=2 the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) --> \frac{2^3}{3}+c*2^2+\frac{2}{6}=5 --> c=\frac{1}{2}. So the formula is: \frac{n^3}{3}+\frac{1}{2}*n^2+\frac{n}{6}. Substitute n=15 to get the sum of the first 15 positive perfect squares: \frac{15^3}{3}+\frac{1}{2}*15^2+\frac{15}{6}=1,240. Answer: C.
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Re: The sum of the first n positive perfect squares, where n is [#permalink]
17 May 2012, 14:55
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carcass wrote: The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?
A. 1,010
B. 1,164
C. 1,240
D. 1,316
E. 1,476 Hello Carcass here it is goes c The formula I have is (n(n+1) (2n+1))/6 n= 15 Hence 15 x16 X31/6 1240 Hope this help best regards
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Re: The sum of the first n positive perfect squares, where n is [#permalink]
17 May 2012, 18:34
Thanks keiraria  Infact I found the same formula (in that way is more simple) in this post sum-of-squares-90497.html. So, I would ask to Bunuel two things: first of all this is a bit unclear: for n=2 the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) . Please, can you elaborate ?? I would like to attack a problem with as methods as possible, because during the exam if you are stuck, can try to find a solution in another way. Secondly, I think this formula is very useful but I didn't find it on Gmat math book why ?? Thanks Bunuel.
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Re: The sum of the first n positive perfect squares, where n is [#permalink]
18 May 2012, 01:09
carcass wrote: Thanks keiraria Infact I found the same formula (in that way is more simple) in this post sum-of-squares-90497.html. So, I would ask to Bunuel two things: first of all this is a bit unclear: for n=2 the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) . Please, can you elaborate ?? I would like to attack a problem with as methods as possible, because during the exam if you are stuck, can try to find a solution in another way. Secondly, I think this formula is very useful but I didn't find it on Gmat math book why ?? Thanks Bunuel. 1. The formula in the stem (n^3/3 + c*n^2 + n/6) gives the sum of the first n positive perfect squares. Notice that the the value of constant c is unknown, so in order to find the sum of the first 15 positive perfect squares we should find its value. We know that the sum of the first two perfect square is 1^2+2^2=5. So, if we substitute n=2 in the formula it should equal to 5: 2^3/3 + c*2^2 + 2/6=5. From here we can find the value of c --> c=1/2 --> formula becomes: \frac{n^3}{3}+\frac{1}{2}*n^2+\frac{n}{6} and now we can substitute n=15 to get the answer. 2. There is a direct formula (given in my post in the earlier thread) to get the sum of the first n positive perfect squares: \frac{N(N + 1)(2N + 1)}{6} --> if n=15 then Sum=\frac{N(N + 1)(2N + 1)}{6}=\frac{15(15 + 1)(2*15 + 1)}{6}=1240. If you know it that's fine but there are thousands of such kind formulas and you certainly cannot and should not memorize them all. For example this formula is not a must know for the GMAT.
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The sum of the first n positive perfect squares, where n is [#permalink]
13 Dec 2012, 00:32
Let n = 1... Sum of 1 perfect squares is 1...
\frac{(1)^3}{3}+c*(1)^2+\frac{1}{6}=1
c=\frac{1}{2}
When n=15:
\frac{(15)^3}{3}+(15)^2/2+\frac{15}{6}=1240
Answer: 1240
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Re: The sum of the first n positive perfect squares, where n is [#permalink]
19 Dec 2012, 07:25
First we need to find the constant 'c'. The easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively.
Hence LHS = 1+4 and plug n=2 for RHS and simplify to get c = 1/2.
Plug values of n = 15 and c = 1/2 into the equation and simplify to get the answer 1240.
Option C.
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Re: The sum of the first n positive perfect squares, where n is
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19 Dec 2012, 07:25
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