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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Find the the sum of the first 20 terms of this series which  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Senior Manager Joined: 20 Feb 2006 Posts: 367 Find the the sum of the first 20 terms of this series which [#permalink] ### Show Tags Updated on: 15 Jun 2010, 11:33 2 15 00:00 Difficulty: 65% (hard) Question Stats: 65% (02:44) correct 35% (02:45) wrong based on 225 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 Show working please! Originally posted by MBAlad on 02 Dec 2006, 19:03. Last edited by alexsr on 15 Jun 2010, 11:33, edited 2 times in total. Senior Manager Joined: 08 Jun 2006 Posts: 324 Location: Washington DC ### Show Tags 02 Dec 2006, 22:05 2 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 Senior Manager Joined: 20 Feb 2006 Posts: 367 ### 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...........????? Senior Manager Joined: 08 Jun 2006 Posts: 324 Location: Washington DC ### Show Tags 03 Dec 2006, 08:21 2 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 Senior Manager Joined: 20 Feb 2006 Posts: 367 ### 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! Manager Joined: 25 Nov 2006 Posts: 58 ### 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. Senior Manager Joined: 20 Feb 2006 Posts: 367 ### 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!!!! Senior Manager Joined: 23 May 2005 Posts: 261 Location: Sing/ HK ### Show Tags 04 Dec 2006, 10:03 1 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 GMAT Club Legend Status: Um... what do you want to know? Joined: 03 Jun 2007 Posts: 5460 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 1 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. Current Student Joined: 28 Dec 2004 Posts: 3251 Location: New York City Schools: Wharton'11 HBS'12 ### Show Tags 17 Jul 2007, 10:21 2 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. Intern Joined: 22 Jun 2009 Posts: 2 Re: Yet Another Sequence! [#permalink] ### 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). Intern Joined: 14 Apr 2010 Posts: 5 Re: Yet Another Sequence! [#permalink] ### Show Tags 14 Apr 2010, 10:17 1 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. Intern Joined: 22 Jun 2010 Posts: 6 Re: Yet Another Sequence! [#permalink] ### 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..... EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12841 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 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
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Find the the sum of the first 20 terms of this series which  [#permalink]

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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.
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Joined: 02 Aug 2009
Posts: 7022
Re: Find the the sum of the first 20 terms of this series which  [#permalink]

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

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12 Aug 2016, 08:14
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Re: Find the the sum of the first 20 terms of this series which  [#permalink]

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21 Sep 2018, 14:26
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

Arithmetic progression )
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Posts: 1110
Find the the sum of the first 20 terms of this series which  [#permalink]

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21 Sep 2018, 19:46
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

if t1+t2=3,
t3+t4=7,
and t5+t6=11,
then we can assume a difference of 4
between consecutive term pairs
thus, t9+t10=19
19+4/2=21=median of series
10 term pairs*21 median=210
A
Find the the sum of the first 20 terms of this series which &nbs [#permalink] 21 Sep 2018, 19:46
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