(a) there is no r. It's not a geometric series
Just the formula shows that the ratio changes from term to term
S1 = 1/4
S2 = 1/4 + 1/28 = 2/7
S3 = 2/7 + 1/70 = 21/70 = 3/10
Looks like Sn = n/(3n+1)
Using induction,
Sk + 1/(3(k+1)-2)(3(k+1)+1)
= Sk + 1/(3k+1)(3k+4)
= k/(3k+1) + 1/(3k+1)(3k+4)
= (k(3k+4)+1)/(3k+1)(3k+4)
= (3k^2+4k+1)/(3k+1)(3k+4)
= (3k+1)(k+1)/(3k+1)(3k+4)
= (k+1)/(3(k+1)+1)
So that's ok.
So, if Sn = n/(3n+1) = 1/3 - 1/3(3n+1)
S∞ = 1/3
Find the sum of the following convergent series:
(a) Ei=1 1 / (3i - 2)(3i + 1)
I know that it equals:
= 1/4 + 1/28 + 1/70 + 1/130 + 1/208
a= 1/4, but I can't figure out what r =
1/4 * 1/7 = 1/28, but from 1/28 to 1/70 and so on it doesn't work anymore.
2 answers
Ok I see, I was just adding the same fractions on every line instead of just adding the answer from the previous. Thanks Steve!
3 answers