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Find the the sum of the first 20 terms of this series which [#permalink]
 02 Dec 2006, 20:03
Question Stats:
58% (04:02) correct
41% (00:51) wrong based on 0 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!
Last edited by alexsr on 15 Jun 2010, 12:33, edited 2 times in total.
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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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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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–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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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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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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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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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) ....
_________________
Impossible is nothing
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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 
Yesterday was a bad day! 
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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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I cant seem to understand the symbols..can someone repost.
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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]
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 (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]
06 Nov 2009, 09:46
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]
23 Nov 2009, 05:49
Great alternative way from anindyat...
never thought about that
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Re: Yet Another Sequence! [#permalink]
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
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.
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(-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]
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 400-330.....
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Re: Yet Another Sequence!
[#permalink]
09 Jul 2010, 21:26
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