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VP
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Find the the sum of the first 20 terms of this series which [#permalink]
 13 Mar 2005, 19:19
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
Last edited by Antmavel on 13 Mar 2005, 20:56, edited 1 time in total.
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GMAT Club Legend
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antmavel, do you meant 1^2 in your question or something else?? The power is missing from your question.
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GMAT Club Legend
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I assume you're missing out a ^2 for base '1'.
The series will work out to be an arithmetic progression with 10 terms, and a common difference of 4. The sum will therefore be 10/2(2(3) + 9(4)) = 210 (A)
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Manager
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sum of squares = n*(n+1).(2n+1) / 6
summing all the +ve nos : 2^2+4^2+6^2...= 4(1^2+...+10^2) = 1540
sum of all -ve nos = (sum of squares upto 20) - 1540 = 1330
Sum asked = 1540 - 1330 = 210.
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SVP
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"A"
series will be:
-1+ 2^2-3^2+4^2....
3,7,11,15,19,23....of 10 terms...of equally spaced numbers with diff of 4
Avg = (19+23)/2 = 21
21 = (SUM)/10
SUM = 210
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VP
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ywilfred wrote: antmavel, do you meant 1^2 in your question or something else?? The power is missing from your question.
oooops  sorry, I've modified it...
you were right, OA is A
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SVP
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A
=-1+4-9+.................. -361+400
=3+7+11+...................+39
a1=3
d=an-a1=7-3=11-7=4
sum of the series = a1+a2+a3+a4+....+a10=210
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SVP
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I'd do it this way:
2^2-1^2=(2+1)(2-1)
4^2-3^2=(4+3)(4-3)
...
Therefore S=1+2+3+4+...19+20=(1+20)*20/2=210
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SVP
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+ nos=> first+last/2=>(4+400)/2*10=>2020
- nos=> (-1+(-361))/2*10=>1810
2020-1810=210
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Senior Manager
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i used the average of the 5th and 6th term and multiplied it by the total number of terms...210
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Director
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HongHu wrote: I'd do it this way:
2^2-1^2=(2+1)(2-1)
4^2-3^2=(4+3)(4-3)
...
Therefore S=1+2+3+4+...19+20=(1+20)*20/2=210
great way to do 
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Senior Manager
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[quote="ywilfred"]I assume you're missing out a ^2 for base '1'.
The series will work out to be an arithmetic progression with 10 terms, and a common difference of 4. The sum will therefore be 10/2(2(3) + 9(4)) = 210 (A)[/quote
I like this formula. Can somebody pls explain?
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Director
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HongHu wrote: I'd do it this way:
2^2-1^2=(2+1)(2-1)
4^2-3^2=(4+3)(4-3)
...
Therefore S=1+2+3+4+...19+20=(1+20)*20/2=210
I was really lost on this one. Thanks HongHu
_________________
Praveen
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Director
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HongHu wrote: I'd do it this way:
2^2-1^2=(2+1)(2-1)
4^2-3^2=(4+3)(4-3)
...
Therefore S=1+2+3+4+...19+20=(1+20)*20/2=210
Elegance par excellence.......is your solution !!!!
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Manager
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I did it like this
Add 2^2 + 4^2 + 6^2... and subtract 2^2 + 4^2 + 6^2...
So the series becomes - (1^2 + 2^2 + 3^2...20^2) + 2 ( 2^2 + 4^2 + 6^2..+20^2)
= - n(n+1)(2n+1)/6 (Here n = 20) + 2*4*n(n+1)(2n+1)/6 (Here n = 10)
= -410*7 + 440*7 = 210
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close
