3
$\begingroup$

Find the following sum

$$ \frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1} $$

I am not able to find any short trick for it. Is there any short trick or do we have to simplify and add it?

$\endgroup$
8
$\begingroup$

$$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$

$\endgroup$
1
$\begingroup$

Lemma to find the sum of $n$ terms of a series each term of which is composed of the reciprocal of the product of $r$ factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression:

Write down the $nth$ term, strike off a factor from the beginning, divide by the number of factors so diminished and by the common difference, change the sign and add a constant.

In our case, the general term is $T_n$=$$\frac 1 {(2n)^2-1}$$

Which is simply, $$\frac 1 {(2n-1)(2n+1)}$$

Applying the lemma, we get:

$$S_n=C-(\frac 1 2)(\frac 1 {2n+1})$$

$$S_n=C-\frac 1 {4n+2}$$ Where $C$ is a constant quantity.

Putting $n=1$ in general term, we get:

$$S_1=\frac 1 3=C- \frac 1 {4+2}$$

That gives $$C=\frac 1 2$$

Thus, $$S_n=\frac 1 2-\frac 1 {4n+2}$$

Putting $n=10$ gives:

$$S_{10}=\frac 1 2-\frac 1 {40+2}$$ $$S_{10}=\frac 1 2-\frac 1 {42}$$ $$Or, S_{10}=\frac {20} {42}=\frac {10} {21}$$
Which is the desired answer.

The method is a bit, long... But it is great for the so-called "bashing" through such problems. If you wish to see a proof, I refer you to Hall and Knight's Higher Algebra.

(P.S. I did this long answer with this specific method just so I could get a hang of the typing system in StackExchange. So, please forgive me even if the method seems useless... I'm new here...)

$\endgroup$
0
$\begingroup$

Alternatively to the telescoping sum decomposition, there is an easy pattern

$$\frac13$$

$$\frac13+\frac1{15}=\frac25$$

$$\frac13+\frac1{15}+\frac1{35}=\frac37$$

$$\frac13+\frac1{15}+\frac1{35}+\frac1{63}=\frac49$$

$$\cdots$$

$\endgroup$
  • $\begingroup$ To use this rigorously in an open-ended answer (as opposed to just multiple choice), one needs to prove the pattern by induction. $\endgroup$ – Deepak Jun 26 at 7:11

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.