sequences

• Apr 1st 2009, 05:23 AM
Webby
sequences
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 AM
stapel
Quote:

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

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 PM
CaptainBlack
Quote:

Originally Posted by stapel
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

The second should be:

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

CB
• Apr 1st 2009, 09:16 PM
Webby
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.
• Apr 1st 2009, 11:44 PM
CaptainBlack
Quote:

Originally Posted by Webby
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.

Could you restructure this more systematically so that we can understand what you have tried to do.

(also, this would only work for n even, you will have to treat n odd seperatley)

CB
• Apr 1st 2009, 11:50 PM
Webby
Quote:

Originally Posted by CaptainBlack
Could you restructure this more systematically so that we can understand what you have tried to do.

(also, this would only work for n even, you will have to treat n odd seperatley)

CB

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