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

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Find the the sum of the first 20 terms of this series which [#permalink] New post 02 Dec 2006, 19:03
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

47% (04:02) correct 53% (01:43) wrong based on 15 sessions
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!
[Reveal] Spoiler: OA

Last edited by alexsr on 15 Jun 2010, 11:33, edited 2 times in total.
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 [#permalink] New post 02 Dec 2006, 22:05
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
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 [#permalink] New post 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...........?????
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 [#permalink] New post 03 Dec 2006, 08:21
–1^2 + 2^2 – 3^2 + 4^2 – 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
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 [#permalink] New post 03 Dec 2006, 09:09
anindyat wrote:
–1^2 + 2^2 – 3^2 + 4^2 – 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!
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 [#permalink] New post 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.
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 [#permalink] New post 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!!!!
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 [#permalink] New post 04 Dec 2006, 10:03
let me try to explain...

work 2 digits by 2 digits

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

next...

– 3^2 + 4^2 = 4^2 – 3^2 = (4-3)(4+3)

Hence you get (1+2) + (3+4) ....
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 [#permalink] New post 04 Dec 2006, 15:32
Hermione wrote:
let me try to explain...

work 2 digits by 2 digits

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

next...

– 3^2 + 4^2 = 4^2 – 3^2 = (4-3)(4+3)

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


Thanks Hermione but I got it late last night :sleep

Yesterday was a bad day! :???
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 [#permalink] New post 16 Jul 2007, 22:41
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.
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 [#permalink] New post 17 Jul 2007, 08:39
I cant seem to understand the symbols..can someone repost.
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 [#permalink] New post 17 Jul 2007, 10:21
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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.
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Re: Yet Another Sequence! [#permalink] New post 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).
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Re: Yet Another Sequence! [#permalink] New post 06 Nov 2009, 08:46
I followed the same way as kryzak's. But I think anindyat's way is a bit faster.
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Re: Yet Another Sequence! [#permalink] New post 23 Nov 2009, 04:49
Great alternative way from anindyat...
never thought about that
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Re: Yet Another Sequence! [#permalink] New post 14 Apr 2010, 10:17
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.
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Re: Yet Another Sequence! [#permalink] New post 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.....
Re: Yet Another Sequence!   [#permalink] 09 Jul 2010, 20:26
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