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
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!
Sorry for taking so long, i have had electricity issues .
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.
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