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The sum of the first n positive perfect squares, where n is

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The sum of the first n positive perfect squares, where n is [#permalink] New post 17 May 2012, 12:43
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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
[Reveal] Spoiler: OA

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Re: What is the sum of the first 15 positive perfect squares? [#permalink] New post 17 May 2012, 13:50
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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] New post 17 May 2012, 13:55
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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] New post 17 May 2012, 17:34
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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] New post 18 May 2012, 00:09
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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] New post 12 Dec 2012, 23:32
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] New post 19 Dec 2012, 06:25
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 [#permalink] New post 16 Jul 2015, 14:12
Is there an easy way to simply the \(15^3/3\) without doing it the multiplying then dividing?
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The sum of the first n positive perfect squares, where n is [#permalink] New post 16 Jul 2015, 15:15
xLUCAJx wrote:
Is there an easy way to simply the \(15^3/3\) without doing it the multiplying then dividing?


Why do you need to calculate \(15^3/3\), when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to \(n^3/3+n^2/2+n/6\) ----> \(\frac{n(n+1)(2n+1)}{6}\)

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating \(15^3/3\)
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Re: The sum of the first n positive perfect squares, where n is [#permalink] New post 16 Jul 2015, 15:42
Engr2012 wrote:
xLUCAJx wrote:
Is there an easy way to simply the \(15^3/3\) without doing it the multiplying then dividing?


Why do you need to calculate \(15^3/3\), when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to \(n^3/3+n^2/2+n/6\) ----> \(\frac{n(n+1)(2n+1)}{6}\)

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating \(15^3/3\)


What rule allows you to do 15*16/6= 5*8?
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Re: The sum of the first n positive perfect squares, where n is [#permalink] New post 16 Jul 2015, 16:20
xLUCAJx wrote:
Engr2012 wrote:
xLUCAJx wrote:
Is there an easy way to simply the \(15^3/3\) without doing it the multiplying then dividing?


Why do you need to calculate \(15^3/3\), when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to \(n^3/3+n^2/2+n/6\) ----> \(\frac{n(n+1)(2n+1)}{6}\)

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating \(15^3/3\)


What rule allows you to do 15*16/6= 5*8?


You can follow the process mentioned below:

15*16/6 ---> cancel factor of 3 out of 15 and 6 to get 5*16/2 ---> now cancel factor of 2 out of 16 and to get ---> 5*8/1---> 5*8

Look at the following reducing-fractions-or-portions-without-making-a-mistake-186285.html#p1423965
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Re: The sum of the first n positive perfect squares, where n is   [#permalink] 16 Jul 2015, 16:20
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