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

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

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13 Mar 2005, 19:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the the sum of the first 20 terms of this series which begins this way : â€“1^2 + 2^2 â€“ 3^2 + 4^2 â€“ 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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13 Mar 2005, 20:18
antmavel, do you meant 1^2 in your question or something else?? The power is missing from your question.
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13 Mar 2005, 20:25
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)
Manager
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13 Mar 2005, 20:33
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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13 Mar 2005, 20:39
"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
VP
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13 Mar 2005, 20:57
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
VP
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13 Mar 2005, 21:46
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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13 Mar 2005, 22:19
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
VP
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14 Mar 2005, 03:17
+ nos=> first+last/2=>(4+400)/2*10=>2020
- nos=> (-1+(-361))/2*10=>1810
2020-1810=210
Manager
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14 Mar 2005, 13:04
i used the average of the 5th and 6th term and multiplied it by the total number of terms...210
Director
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14 Mar 2005, 14:49
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
Senior Manager
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19 Mar 2005, 14:08
[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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20 Mar 2005, 01:20
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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20 Mar 2005, 06:27
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 !!!!
Manager
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20 Mar 2005, 16:46
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
20 Mar 2005, 16:46
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