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

Question

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), 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

Show Answer

Official Answer

C

(A) 1,010
(B) 1,164
(C) 1,240
(D) 1,316
(E) 1,476

gurpreetsingh · Accepted Answer

IMO C

1 is a perfect square put n = 1

sum is  1 = \frac{1}{3} + \frac{1}{6} + c

=>  c = \frac{1}{2} , now put n= 15 and  c = \frac{1}{2}  in  \frac{n^3}{3} + cn^2 + \frac{n}{6}

This gives sum = 1240

Bunuel · Answer

The formula for counting the sum of the first  n  positive perfect squares is:  \frac{N(N + 1)(2N + 1)}{6} .

n=15 ,  Sum=1240

Answer: C.

srini123 · Answer

I didnt get the formula "n3/3 + cn2 + n/6", can you use {math function} to post , so we an read it correctly.

My approach would be to try  1^2  or  2^2  and find the value of c and then use n=15 to find the answer

AtifS · Answer

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), 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

Show Answer

Official Answer

C

(A) 1,010
(B) 1,164
(C) 1,240
(D) 1,316
(E) 1,476

Appeared on mgmat site as Problem challenge of the week..
I used the following approach to solve it but it isn't equal to any of the choices.
by using the formula we may get
Let's take the constant c=1
15^3/3 + 15^2 +15/6=15(15^2/3+15+1/6)=15(225/3+15+1/6)=15(75+15+1/6)=15(80+1/6)=15(481/6)=(5*481)/2=2405/2=1202.5

Any thoughts upon this question? Also don't know the OA

AtifS · Answer

@gurpreetsingh...yea I also chose C. but if we put c=1/2 then the answer would be 1195 correct me if I am wrong

gurpreetsingh · Answer

sum =  \frac{( 15*15*15)}{3} + \frac{15*15}{2}+ \frac{15}{6} =  \frac{15*15*10}{2} + \frac{15*15}{2} + \frac{5}{2}

=  \frac{(2250+225+5)}{2} 
=  \frac{2480}{2} 
= 1240

On what basis you took c =1? 
Try to analyze.

nonameee · Answer

What is a perfect square and is this concept tested on the GMAT? Thank you.

Bunuel · Answer

A perfect square, is just an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Check Number Theory chapter of Math Book for more: math-number-theory-88376-60.html

Bunuel · Answer

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.

Carcass · Answer

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.

Bunuel · Answer

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.

mbaiseasy · Answer

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

EMPOWERgmatRichC · Answer

Hi ALL,

Certain Quant questions on the GMAT require a bit of "playing around"; You can figure out the value of C (the unknown constant in the question) by TESTing Values - TEST any number of perfect squares, sum them up and you can solve for C. With that value in place, you can then follow the rest of the prompt, plug in N=15 and solve.

It's important to remember that most questions CAN be solved in more than one way, so keep an open mind as to what else you might do to answer a given question. Here, we actually have a question that can also be solved without the given formula. The process would take a bit longer, but there are a couple of pattern-matching shortcuts that would save some serious time.

First, name the first 15 positive perfect squares (this is knowledge that you should have memorized before Test Day): 
1 
4 
9 
16 
25

36 
49 
64 
81 
100

121 
144 
169 
196 
225

Don't add them up just yet. Notice that the "unit's digit" of the ANSWERS are limited to 3 options (0, 4 or 6). Now, let's pair off those 15 numbers….

1 + 9 = ends in a 0 
4 + 16 = ends in a 0 
Notice a pattern….. 
Every group of 5 values has a "1" and a "9" and a "4" and a "6", which means that we have lots of numbers that sum to a unit's digit of 0.

All that's left are the multiples of five: 25, 100 and 225….those numbers, when summed, end in a 0.

This means that the total sum must also end in a 0. Eliminate B, D and E.

Now, let's estimate, starting with the larger values and working to the smaller ones…. 
225 + 196 = over 400 
169 + 144 = over 300 
121 + 100 + 81 = over 300 
Total so far = over 1,000 
And we have all those other values to add in….The total MUST be greater than 1010. Eliminate A.

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

WiziusCareers1 · Answer

The sum of the first 2 perfect squares (1 and 4) is 5.
From the formula, we get (n^3)/3 + cn^2 + n/6 = 5
Putting n = 2, we get 8/3 + 4n + 1/3 = 5
=> c = 0.5

So, the sum of the first 15 positive perfect squares is (15^3)/3 + (0.5)(15^2) + 15/6 = 1125 + 112.5 + 2.5 = 1240

The answer is (C).

MathRevolution · Answer

The sum of the first n positive perfect squares:  \frac{n (n + 1) (2n + 1) }{ 6}

For n = 15 :  \frac{n (n + 1) (2n + 1) }{ 6} = \frac{15 * 16 * 31 }{ 6} = 5 * 8 * 31 = 40 * 31 = 1,240

Answer C

luisdicampo · Answer

Deconstructing the Question  
The formula for the sum of the first  n  perfect squares is:
 \frac{n^3}{3}+c n^2+\frac{n}{6}

We first determine the constant  c  by plugging in a small value of  n .

Step-by-step  
For  n=1 , the sum is  1 .

\frac{1^3}{3}+c(1)^2+\frac{1}{6}=1

\frac{1}{3}+c+\frac{1}{6}=1

\frac{1}{2}+c=1

c=\frac{1}{2}

Now the formula becomes:

\frac{n^3}{3}+\frac{1}{2}n^2+\frac{n}{6}

Combine over denominator 6:

\frac{2n^3+3n^2+n}{6}=\frac{n(2n^2+3n+1)}{6}

Substitute  n=15 :

\frac{15(2(225)+45+1)}{6}

\frac{15(496)}{6}

=1240

Answer: C

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

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

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