0
$\begingroup$

I'm reading the resolution of a solved exercise about Telescoping Series in a "Finite Math" course I'm attending. The resolution has a step that is not explained because it's left as an exercise to the reader:


$$ \cdots = \frac{1}{(2k + 3)(2(k + 1) +3)} = \frac{1}{2} \bigg(\frac{1}{2k + 3} - \frac{1}{2(k + 1) + 3}\bigg)$$


I'm guessing the missing step(s) - that turned the fraction into the last "half a difference" of fractions - uses "Partial Fraction Decomposition" technique(s) (that is also part of the Syllabus of this course), but I'm not sure about that and I'm not yet familiar with that technique(s).

So, I read the Wikipedia article about "Partial Fraction Decomposition", namely the "Example 1" section and tried to apply it here, but I get "stuck" quickly:

$$\frac{1}{(2k + 3)(2(k + 1) +3)} = \frac{A}{2k + 3} + \frac{B}{2(k + 1) + 3} \Leftrightarrow$$

$$\Leftrightarrow A[2(k + 1)+3] + B(2k +3) = 1$$

... and I don't know how to carry on from here.

Am I right when I think this is solved using "Partial Fraction Decomposition"? Have I begun solving this well? What should I do now?

$\endgroup$
1
$\begingroup$

This breaks down into
$(A2+B2)k+ (5A+3B) = 1$ or since k is an indeterminate...
$(A2+B2)k+ (5A+3B) = 1 + 0k$ So from there you simply match up the coefficients. $(A2+B2) = 0 $ & $(5A+3B)=1$ You can then solve for A and B.

$\endgroup$
  • $\begingroup$ @wfwood Thanks! :-) $\endgroup$ – ricmarques Sep 1 '13 at 18:08
2
$\begingroup$

You want to cross multiply the fractions and then equate both sides.

We have:

$$1 = A(2k + 5) + B(2k+3)$$

This leads to two equations with two unknowns.

  • $A+B = 0$
  • $5A + 3B = 1$.

Solving these leads to $A = \dfrac{1}{2}, B = -\dfrac{1}{2}$.

You might also want to look up Partial Fraction Cover Up Method and using limits to solve for the constants.

$\endgroup$
  • 1
    $\begingroup$ Nice work! $\langle +1\rangle$ $\endgroup$ – Namaste Sep 1 '13 at 18:02
  • $\begingroup$ @Amzoti Thanks for the answer and for the reference to "Partial Fraction Cover Up Method" :-) I upvoted your reply but ended up accepting the answer of "wfwood" because it was similar, a minute earlier than yours and - last but not least - "wfwood" seems to "need" the extra reputation a bit more than you! ;-) I would like to accept both answers, but I can't :-( $\endgroup$ – ricmarques Sep 1 '13 at 18:07
  • 1
    $\begingroup$ @ricmarques: You are welcome and no worries, the goal is to help people and I wish we could hide all of our reputation, because the best answers should rise to the top! :-) Regards $\endgroup$ – Amzoti Sep 1 '13 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.