May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 20 Feb 2006
Posts: 362

Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
Updated on: 15 Jun 2010, 12:33
Question Stats:
65% (02:45) correct 35% (02:44) wrong based on 232 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!
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by MBAlad on 02 Dec 2006, 20:03.
Last edited by alexsr on 15 Jun 2010, 12:33, edited 2 times in total.



Senior Manager
Joined: 08 Jun 2006
Posts: 314
Location: Washington DC

Getting a
this is pair of a^2  b^2
after breaking it in (a+b)(ab)
Sum = (1+2) + (3+4) +...(19+20)
Sum = 210



Senior Manager
Joined: 20 Feb 2006
Posts: 362

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: 314
Location: Washington DC

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)(2019) + (18+17)(1817)+......+ (2+1)(21)
= 20 + 19 + 18 + 17.....+ 3



Senior Manager
Joined: 20 Feb 2006
Posts: 362

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)(2019) + (18+17)(1817)+......+ (2+1)(21) = 20 + 19 + 18 + 17.....+ 3
I'm half way there:
How does (20+19)(2019) + (18+17)(1817)+...... 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

take the first and last digit,
on applying (a^2b^2) formula we get
21(19)21(17)+21(15)21(13)+...............21(1)
21(1917+1513+119+75+31)
21(2+2+2+2+2)
21(10)=210.



Senior Manager
Joined: 20 Feb 2006
Posts: 362

Hi pzazz
Quote: take the first and last digit, on applying (a^2b^2) formula we get 21(19)21(17)+21(15)21(13)+...............21(1) 21(1917+1513+119+75+31) 21(2+2+2+2+2) 21(10)=210.
I dont see how you've applied the (a^2b^2) formula?
21(19) = 399?
Also why did you use 21?????
I will get there!!!!



Senior Manager
Joined: 23 May 2005
Posts: 260
Location: Sing/ HK

let me try to explain...
work 2 digits by 2 digits
1^2 +2^2 = 2^2 1^2 = (2+1)(21)
next...
â€“ 3^2 + 4^2 = 4^2 â€“ 3^2 = (43)(4+3)
Hence you get (1+2) + (3+4) ....
_________________



GMAT Club Legend
Status: Um... what do you want to know?
Joined: 04 Jun 2007
Posts: 5460
Location: SF, CA, USA
Schools: UC Berkeley Haas School of Business MBA 2010
WE 1: Social Gaming

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 2^21^2 = 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) + (101)(4)]/2 = 210.



Current Student
Joined: 28 Dec 2004
Posts: 3157
Location: New York City
Schools: Wharton'11 HBS'12

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
(ab)(a+b) here b =1 and a=2;
(21)(2+1)+(43)(4+3)...basically it becomes 1+2+3....+20
so the sum of the first 20 integers is (20+1)/2 * (201)+1=210
A it is.



Intern
Joined: 22 Jun 2009
Posts: 2

Re: Yet Another Sequence!
[#permalink]
Show Tags
29 Jun 2009, 14:13
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 (21)(2+1) + (43)(4+3) + (65)(6+5) + (87)(8+7) + ..... + (2019)(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, 11:17
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, 21:26
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 400330.....



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14198
Location: United States (CA)

Re: Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
06 Jun 2015, 16:49
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*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



Board of Directors
Joined: 17 Jul 2014
Posts: 2552
Location: United States (IL)
Concentration: Finance, Economics
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, 18:58
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 400361 = 39 324289= 35 256225=31 196169=27 144121=23 10081=19 6449=15 3625=9 169=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.



Math Expert
Joined: 02 Aug 2009
Posts: 7685

Re: Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
03 Apr 2016, 03:08
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 400361 = 39 324289= 35 256225=31 196169=27 144121=23 10081=19 6449=15 3625=9 169=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
_________________



Current Student
Joined: 12 Aug 2015
Posts: 2617

Re: Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
12 Aug 2016, 09:14
Magnificent Question Here using the a^2b^2 property => it literally reduces to 1+2+3+4+5+6+.....20=> Sum =210
_________________



Senior Manager
Joined: 26 Jun 2017
Posts: 408
Location: Russian Federation
Concentration: General Management, Strategy
WE: Information Technology (Other)

Re: Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
21 Sep 2018, 15:26
MBAlad wrote: 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! Arithmetic progression )



VP
Joined: 07 Dec 2014
Posts: 1192

Find the the sum of the first 20 terms of this series which
[#permalink]
Show Tags
21 Sep 2018, 20:46
MBAlad wrote: 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! 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
[#permalink]
21 Sep 2018, 20:46






