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April 17, 2015

April 17, 2015

Posted by **Sean** on Wednesday, March 18, 2009 at 12:44am.

Sigma (k = 1 to infinity): 1/(9k^2 + 3k - 2)

- Math -
**Reiny**, Wednesday, March 18, 2009 at 7:49amterm(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

- Math -
**Sean**, Wednesday, March 18, 2009 at 11:16amThanks 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

- Math -
**Reiny**, Wednesday, March 18, 2009 at 12:08pmfor 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

- Math -
**Reiny**, Wednesday, March 18, 2009 at 12:14pmAlso 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 -
**Anonymous**, Wednesday, March 18, 2009 at 1:37pmjko[j

- Math -
**deirra**, Wednesday, March 18, 2009 at 4:35pm5_13+7_61=11_20

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