term(k) = 1/[(3k+2)(3k-1)]
so S(1) = 1/(5 x 2) = 1/10
S(2) = 1/10 + 1/(8 x 5) = 1/8 or 2/16
S(3) = 1/8 + 1/88 = 12/88 = 3/22
S(4) = 3/22 + 1/154 = 1/7 or 4/28
it appears we have a nice pattern here and
S(n) = n/(6n+4)
This type of question usually comes up with the topic of "induction".
You would now have to prove that this conjecture is true by induction
Thanks Reiny! I couldn't factor that polynomial. Once there, the rest follows the book example
Each term in the sequence: 1/(9k^2 + 3k - 2) = 1/((3k+2)(3k-1)) = 1/3*(1/(3k-1) - 1/(3k+2))
When you take a sum of those terms, it's a telescoping series where the -1/(3k+2) cancels the next +1/(3k-1) term.
The sum of a partial series from terms 1 to n = 1/3*(1/2 - 1/(3n + 2)). Limit to infinity = 1/6
for your additional information:
You might want to remember that in these kind of telescoping series, the two factors are of this pattern:
(mk + a)(mk + b) where a - b = m
in our case they were (3k+2)(3k-1)
notice 2-(-1) = 3
Also notice that if you simplify your
1/3*(1/2 - 1/(3n + 2)) you get my n/(6n+4)
and Limit n/(6n+4) as n ---> ∞ = 1/6
Math - What does the following infinite series starting at k=2 converge to: &...
how would I.... - how would I express the repeating decimal representaion of 1/9...
CALCULUS-URGENT- no one will respond!!! - we know the series from n=0 to ...
Calculus-Series - Sigma with n=1 to n= positive infinity (x^3)* (e^(-x^4)) Does ...
CALCULUS-URGENT - we know the series from n=0 to infinity of c(sub n)*3^n ...
Calculus - determine whether the series converges or diverges. I am stuck on ...
math - I read from my textbook: If S is the infinite series 1 + x + x^2 + x^3...
math - 1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4 A)15 B)15/2...
CALCULUS - we know the series from n=0 to infinity of c(sub n)*3^n converges 1...
Calc II - Use the comparison or limit comparison test to decide if the following...