Find the sum to n terms of the sequence 100-2^2 +108-4^2 +116-6^2+ ...

All help is much appreciated. Thank you

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- Apr 1st 2009, 05:23 AMWebbysequences
Find the sum to n terms of the sequence 100-2^2 +108-4^2 +116-6^2+ ...

All help is much appreciated. Thank you - Apr 1st 2009, 05:58 AMstapel
This appears to be the subtraction of two different series:

. . . . .$\displaystyle 100\, +\, 108\, +\, 116\, +\, ...\, =\, \sum_{i=0}^{n-1}\, 100\, +\, 8i$

. . . . .$\displaystyle 2^2\, +\, 4^2\, +\, 6^2\, +\, ...\, =\, \sum_{i=0}^{n-1}\, \left(2(i\, +\, 2)\right)^2$

What formulas do you have for these sorts of series?

When you reply, please include a clear listing of your thoughts and working so far. Thank you! :D - Apr 1st 2009, 08:38 PMCaptainBlack
- Apr 1st 2009, 09:16 PMWebby
Sorry for taking so long, i have had electricity issues (Worried).

To find the sum to n terms for a sequence such as this i have been using sigma notation.

My current workings;

100 + 108 + 116 +... to n/2 terms, -(2^2+4^2+6^2+...) to n/2 terms

(8n+92) -(2n)^2

8(n/2 sigma r=1)r + (n/2 sigma r=1)92 - 4(n/2 sigma r=1)r^2

n(n+2) + 46n - n(n+2)(n+1)/6

Substituting a values for n into this equation is not giving me the sum to n terms. Hence i have done something wrong up to this point.

If you cannot understand anything please ask. - Apr 1st 2009, 11:44 PMCaptainBlack
- Apr 1st 2009, 11:50 PMWebby
If instead of having it to n/2 terms, i had it to n terms. Would that make the sum to n terms work for all real numbers? This could be my mistake.

**My current workings**;

100 + 108 + 116 +... to n/2 terms, -(2^2+4^2+6^2+...) to n/2 terms

(8n+92) -(2n)^2

8(n/2 sigma r=1)r + (n/2 sigma r=1)92 - 4(n/2 sigma r=1)r^2

using r = n(n+1)/2 r^2 = n(n+1)(2n+1)/6

8 * (n/2)((n/2)+1)/2 + 92n/2 + 4 * (n/2)((n/2)+1)((2n/2)+1)/6

n(n+2) + 46n - n(n+2)(n+1)/6