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

Printable View

- Apr 1st 2009, 06:23 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, 06:58 AMstapel
- Apr 1st 2009, 09:38 PMCaptainBlack
- Apr 1st 2009, 10:16 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 2nd 2009, 12:44 AMCaptainBlack
- Apr 2nd 2009, 12:50 AMWebby
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