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sarah12345
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#1
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#1
find the sum of (n+1)^2 + (n+2)^2+...........+(2n)^2

this is a worked example question and it says

notice that this is the sum 1^2 + 2^2 +......+n^2+.....+(2n)^2

with the first n terms subtracted


im being really thick and dont see how they got that could someone explain?

thanks alot
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john !!
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1^2 + 2^2 + 3^2 + 4^2 + ... + (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 + ... + (2n-1)^2 + (2n)^2

the terms underlined are the ones subtracted. all the terms above makes up the sum (of r^2) from r=1 to r=2n. the leftover sum is the sum (of r^2) from r=(n+1) to r=2n.
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lgs98jonee
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Formula for sum of squares from 1 to 2n is:

1/6 . 2n . (4n+1) . (2n+1)

and from n+1 to 2n is

1/6 . 2n . (4n+1) . (2n+1) - 1/6 . n . (2n+1) . (n+1)

=n/6 . (2(4n+1)(2n+1)-(2n+1)(n+1)
=n(2n+1)/6 . (8n+2-n-1)
=1/6. (n(2n+1)(7n+1))

i think.....
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