# summing series

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Thread starter 15 years ago
#1
find the sum of the series

1^2 -2^2 + 3^2 - 4^2+.....+(2n-1)^2 - (2n)^2

i said that is was equal to sigma(between 1 and 2n) r^2 - sigma(between 1 and n) 2r^2

but the the answer is

sigma (between 1 and 2n) r^2 - 8 * sigma(between 1 and n) r^"

I really dont understand where the *8 came from could someone explain where ive gone wrong?

thanks
0
15 years ago
#2
Think of it as

1^2 - 2^2 + ... +(2n-1)^2 - (2n)^2

= [1^2 + 2^2 + 3^2 + ... + (2n)^2] - 2 [2^2 + 4^2 + 6^2 + ... (2n)^2]

= [1^2 + 2^2 + 3^2 + ... + (2n)^2] - 8 [1^2 + 2^2 + 3^2 + ... n^2]

as

(2k)^2 = 4 (k^2).
0
15 years ago
#3
I got it simplified to something like 4n^2 - n or something... dunno if that's right, I lost my message
0
15 years ago
#4
1^2 - 2^2 + 3^2 - 4^2 +... + (2n - 1)^2 - (2n)^2
is equal to [1^2 + 2^2 + 3^2 + 4^2 +... + (2n - 1)^2 + (2n)^2 ] - 2*[ 2^2 + 4^2 +... + (2n)^2 ]
It means that Sum from 1 to 2n of r^2 subtracts Sum from 1 to n of 2*(2r)^2
Therefore, the equation is:
Sigma(1 to 2n) r^2 - 8*Sigma(1 to n) r^2
0
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