It is currently 20 Jan 2018, 03:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Find the the sum of the first 20 terms of this series which

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 20 Feb 2006
Posts: 372

Kudos [?]: 40 [2], given: 0

Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

02 Dec 2006, 19:03
2
KUDOS
12
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

65% (04:20) correct 35% (02:07) wrong based on 178 sessions

### HideShow timer Statistics

Find the the sum of the first 20 terms of this series which begins this way : -1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2...

(a) 210
(b) 330
(c) 519
(d) 720
(e) 190

[Reveal] Spoiler: OA

Last edited by alexsr on 15 Jun 2010, 11:33, edited 2 times in total.

Kudos [?]: 40 [2], given: 0

Senior Manager
Joined: 08 Jun 2006
Posts: 335

Kudos [?]: 67 [1], given: 0

Location: Washington DC

### Show Tags

02 Dec 2006, 22:05
1
KUDOS
Getting a

this is pair of a^2 - b^2
after breaking it in (a+b)(a-b)

Sum = (1+2) + (3+4) +...(19+20)
Sum = 210

Kudos [?]: 67 [1], given: 0

Senior Manager
Joined: 20 Feb 2006
Posts: 372

Kudos [?]: 40 [0], given: 0

### Show Tags

03 Dec 2006, 06:50
I'm missing something here -1^2 = 1, 2^2 = 4 -3^2 = 9

So I'm getting 1 + 4 + 9...........?????

Kudos [?]: 40 [0], given: 0

Senior Manager
Joined: 08 Jun 2006
Posts: 335

Kudos [?]: 67 [1], given: 0

Location: Washington DC

### Show Tags

03 Dec 2006, 08:21
1
KUDOS
â€“1^2 + 2^2 â€“ 3^2 + 4^2 â€“ 5^2 + 6^2............-19^2 + 20^2
= (20^2 -19^2) + (18^2 -17^2)+...(2^2 -1^2)
= (20 + 19)(20-19) + (18+17)(18-17)+......+ (2+1)(2-1)
= 20 + 19 + 18 + 17.....+ 3

Kudos [?]: 67 [1], given: 0

Senior Manager
Joined: 20 Feb 2006
Posts: 372

Kudos [?]: 40 [0], given: 0

### Show Tags

03 Dec 2006, 09:09
anindyat wrote:
â€“1^2 + 2^2 â€“ 3^2 + 4^2 â€“ 5^2 + 6^2............-19^2 + 20^2
= (20^2 -19^2) + (18^2 -17^2)+...(2^2 -1^2)
= (20 + 19)(20-19) + (18+17)(18-17)+......+ (2+1)(2-1)
= 20 + 19 + 18 + 17.....+ 3

I'm half way there:

How does (20+19)(20-19) + (18+17)(18-17)+...... turn into 20+19+18.....??

How did you arrive at 210

Urggghhh I'm having a bad day!

Kudos [?]: 40 [0], given: 0

Manager
Joined: 25 Nov 2006
Posts: 59

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Dec 2006, 09:31
take the first and last digit,
on applying (a^2-b^2) formula we get
21(19)-21(17)+21(15)-21(13)+...............-21(1)
21(19-17+15-13+11-9+7-5+3-1)
21(2+2+2+2+2)
21(10)=210.

Kudos [?]: 2 [0], given: 0

Senior Manager
Joined: 20 Feb 2006
Posts: 372

Kudos [?]: 40 [0], given: 0

### Show Tags

03 Dec 2006, 15:26
Hi pzazz

Quote:
take the first and last digit,
on applying (a^2-b^2) formula we get
21(19)-21(17)+21(15)-21(13)+...............-21(1)
21(19-17+15-13+11-9+7-5+3-1)
21(2+2+2+2+2)
21(10)=210.

I dont see how you've applied the (a^2-b^2) formula?

21(19) = 399?

Also why did you use 21?????

I will get there!!!!

Kudos [?]: 40 [0], given: 0

Senior Manager
Joined: 23 May 2005
Posts: 263

Kudos [?]: 60 [1], given: 0

Location: Sing/ HK

### Show Tags

04 Dec 2006, 10:03
1
KUDOS
let me try to explain...

work 2 digits by 2 digits

-1^2 +2^2 = 2^2 -1^2 = (2+1)(2-1)

next...

â€“ 3^2 + 4^2 = 4^2 â€“ 3^2 = (4-3)(4+3)

Hence you get (1+2) + (3+4) ....
_________________

Impossible is nothing

Kudos [?]: 60 [1], given: 0

GMAT Club Legend
Status: Um... what do you want to know?
Joined: 03 Jun 2007
Posts: 5464

Kudos [?]: 410 [1], given: 14

Location: SF, CA, USA
Schools: UC Berkeley Haas School of Business MBA 2010
WE 1: Social Gaming

### Show Tags

16 Jul 2007, 22:41
1
KUDOS
Sorry for reviving an old thread, but isn't there an easier way to do this since it's related to arithmetic series?

We know that 2^2-1^2 = 3
4^2 - 3^2 = 7
6^2 - 5^2 = 11
and so forth.

So you have a series of 10 numbers, starting with a1 = 3, d = 4, and n = 10.

Sum of this series is [10( 2(3) + (10-1)(4)]/2 = 210.

Kudos [?]: 410 [1], given: 14

Current Student
Joined: 28 Dec 2004
Posts: 3343

Kudos [?]: 328 [2], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

17 Jul 2007, 10:21
2
KUDOS
OK read it again..and the symbol is just a (-)

so we have -1^2 +2^2 - 3^2+4^2....

can be simplifyed..to

(a-b)(a+b) here b =1 and a=2;

(2-1)(2+1)+(4-3)(4+3)...basically it becomes 1+2+3....+20

so the sum of the first 20 integers is (20+1)/2 * (20-1)+1=210

A it is.

Kudos [?]: 328 [2], given: 2

Intern
Joined: 22 Jun 2009
Posts: 2

Kudos [?]: [0], given: 0

### Show Tags

29 Jun 2009, 13:13
Thanks for reposting the question. I couldnt make sense of the symbol either. I am new to GMATClub and I must say that this group is more than just impressive. I appreciate the camaraderie the members show to help each other get their best, and hope to contribute whenever I can. In that spirit....

I got the (2-1)(2+1) + (4-3)(4+3) + (6-5)(6+5) + (8-7)(8+7) + ..... + (20-19)(20+19).... part
But then this translates to 3 + 7 + 11 + 15 +...... + 39, which I solved as 10 terms with next term incremented by 4.
So their sum would be the middle term (average of 19 and 23 = 21) multiplied by 10 = 210.

Using "Sum of n terms of an equally spaced series = middle terms * n" (if n = even, middle term is average of the two middle terms).

Kudos [?]: [0], given: 0

Intern
Joined: 14 Apr 2010
Posts: 5

Kudos [?]: 3 [0], given: 0

### Show Tags

14 Apr 2010, 10:17
1
This post was
BOOKMARKED
I believe this has already been said in other ways, but I got lost in some of the terminology and symbols, so I had to to break it down a little more for myself. Hopefully this helps anyone still confused by this.

This is a arithmetic sequence, but only if we look at it in pairs. So (-1^2 + 2^2) is the first pair. (-3^2 + 4^2) is the second pair and so on. The sum in each pair adds four to the previous pair each time as follows:

(-1^2 + 2^2) = 3 (note: 3 is our starting number)
(-3^2 + 4^2) = 7
(-5^2 + 6^2) = 11
(-7^2 + 8^2) = 15
.
.
.
(-19^2 + 20^2) = 39

As soon as we recognize the sequence, we can use the sum of n terms arithmetic progression formula:
Sum of n terms = (n/2) x (value of 1st term + value of last term)
substitute:
sum of n terms = (10/2) x (3+39) = 5 x 42 = 210

Note: we used 10 as the N # of terms because we turned the 20 original terms into 10 pairs.

For those concerned about speed, on this problem all that we have to do is calculate the first couple of pairs until we see the pattern, then calculate the last pair (-19^2 + 20^2), add it to the outcome of the first pair (-1^2 + 2^2) and multiply by 5. This can be done in way under 2 minutes.

Hope that helps.

Kudos [?]: 3 [0], given: 0

Intern
Joined: 22 Jun 2010
Posts: 9

Kudos [?]: 1 [0], given: 0

### Show Tags

09 Jul 2010, 20:26
if you didn't see that grouping you could always after doing a couple of numbers see that after every addition step which happens to coincide with even numbers there is an overall increase but the value never attains the value of the squared term added...

so you know that after adding 20^2 the answer will be less than 400

at this point, either guess or do a few terms and see that the negative value build up to higher than 70.....ie 400-330.....

Kudos [?]: 1 [0], given: 0

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10712

Kudos [?]: 3778 [0], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

06 Jun 2015, 15:49
Hi sagarag,

This question has a great "visual component" to it, so I'm going to give you some hints and let you try this question again...

First, let's deal with -1^2 + 2^2

1) 2^2 is the equivalent of a 2x2 square. Draw it and include the 4 individual boxes.
2) -1^2 = -1; Draw a line through one of the 4 squares you just drew. You now have 3 squares left. Notice the pattern in the drawing....

3) Try these same steps again with -3^2 + 4^2; you should end up with a larger drawing but the SAME pattern. How many squares are left here?

4) Can you figure out how many squares would be left with -5^2 + 6^2 WITHOUT drawing the picture this time....? And what about the other 'pairs' up values up to -19^2 + 20^2?

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

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3778 [0], given: 173

Board of Directors
Joined: 17 Jul 2014
Posts: 2719

Kudos [?]: 463 [0], given: 211

Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

02 Apr 2016, 17:58
i got to A..but my main concern, does PEMDAS apply here at all or not?
if not, it needs to be specified so.
PEMDAS - parenthesis, exponents, multiplication, division, addition, subtraction.
exponents come first, so any NEGATIVE integer squared is a positive number...no?
my approach, though lengthy:
list all squares (good to remember squares of the first 20 integers):
1 is positive
4 positive
9 negative
16 positive
25negative
36positive
49negative
64positive
81negative
100positive
121negative
144positive
169negative
196positive
225negative
256positive
289negative
324positive
361negative
400positive

now
400-361 = 39
324-289= 35
256-225=31
196-169=27
144-121=23
100-81=19
64-49=15
36-25=9
16-9=7
and +14=5
now 39+35+31+27+23+19+15+9+7+5
group to be easier:
35+15=50
27+23=50
31+39=70
19+9=28
7+5=12

last digit is 0, so C is out
50+50+70+40=210
A

p.s. surprisingly or not, the sum of all numbers between 1 and 20, inclusive is 20(21)/2 = 210.

Kudos [?]: 463 [0], given: 211

Math Expert
Joined: 02 Aug 2009
Posts: 5534

Kudos [?]: 6439 [0], given: 122

Re: Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

03 Apr 2016, 02:08
mvictor wrote:
i got to A..but my main concern, does PEMDAS apply here at all or not?
if not, it needs to be specified so.
PEMDAS - parenthesis, exponents, multiplication, division, addition, subtraction.
exponents come first, so any NEGATIVE integer squared is a positive number...no?
my approach, though lengthy:
list all squares (good to remember squares of the first 20 integers):
1 is positive
4 positive
9 negative
16 positive
25negative
36positive
49negative
64positive
81negative
100positive
121negative
144positive
169negative
196positive
225negative
256positive
289negative
324positive
361negative
400positive

now
400-361 = 39
324-289= 35
256-225=31
196-169=27
144-121=23
100-81=19
64-49=15
36-25=9
16-9=7
and +14=5
now 39+35+31+27+23+19+15+9+7+5
group to be easier:
35+15=50
27+23=50
31+39=70
19+9=28
7+5=12

last digit is 0, so C is out
50+50+70+40=210
A

p.s. surprisingly or not, the sum of all numbers between 1 and 20, inclusive is 20(21)/2 = 210.

Hi,
Quote:
exponents come first, so any NEGATIVE integer squared is a positive number...no?

Yes exponents come first...
But -3^2 is not SQUARE of -IVE integer, it is -IVE of square of positive integer..
so you will first square it and then add -ive sign to it..... $$-3^2= -9$$
If it were (-3)^2, this is what would have been the case given by you.. $$(-3)^2 = 9$$

Quote:
p.s. surprisingly or not, the sum of all numbers between 1 and 20, inclusive is 20(21)/2 = 210.

If you simplify $$-1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2... +20^2$$, you will get 1+2+3+4..+20
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Kudos [?]: 6439 [0], given: 122

Retired Moderator
Joined: 12 Aug 2015
Posts: 2340

Kudos [?]: 998 [0], given: 682

GRE 1: 323 Q169 V154
Re: Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

12 Aug 2016, 08:14
Magnificent Question
Here using the a^2-b^2 property =>
it literally reduces to 1+2+3+4+5+6+.....20=> Sum =210
_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 998 [0], given: 682

Non-Human User
Joined: 09 Sep 2013
Posts: 14223

Kudos [?]: 291 [0], given: 0

Re: Find the the sum of the first 20 terms of this series which [#permalink]

### Show Tags

21 Dec 2017, 23:51
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 291 [0], given: 0

Re: Find the the sum of the first 20 terms of this series which   [#permalink] 21 Dec 2017, 23:51
Display posts from previous: Sort by

# Find the the sum of the first 20 terms of this series which

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.