Using Partial Fractions to Simplify Summations

[Natasha1]

I don't know how the latex coding works sorry...

I have to use the partial fraction technique on 1/(4k^2 - 1)...
So far so good and I get 1 / 2(2k-1) - 1 / 2(2k+1), is this correct?


I now need to show that if E is the sum of terms where k=1 and n is the unknow.

E 1 / 4k^2 - 1 = n / 2n + 1

Please help

 

[verty]

For the first part, notice: 4k^2 - 1 = (2k-1)(2k+1)

I don't understand the second part, it makes no sense. If it was asking for the value of 'n' it would make sense. There's no reason to ask whether E can be expressed in that form when k = 1, I can't see why a teacher would want to ask it.

Also, what does this have to do with sequences/series?

Edit: I see about the sigma notation now, totally misunderstood the question the first time, don't know how to solve it though.

 

[HallsofIvy]

I don't know how the latex coding works sorry...
I have to use the partial fraction technique on 1/(4k^2 - 1)...
So far so good and I get 1 / 2(2k-1) - 1 / 2(2k+1), is this correct?
I now need to show that if E is the sum of terms where k=1 and n is the unknow.
E 1 / 4k^2 - 1 = n / 2n + 1
Please help

$$\Sigma_{k=1}^n\frac{1}{4k^2-1}= \frac{n}{2n+1}$$
Is that what you mean? Even if you can't use latex, please use parentheses! Most people would read 1/4k^2-1 as (1/4)k^2- 1 or 1/(4k^2)- 1 rather than 1/(4k^2-1).
Yes, your partial fraction decomposition is correct. To see how it applies to the sum, write out a few of the terms using that decomposition:
doing, say k= 1 to 4:
k= 1 $$\frac{1}{4-1}= \frac{1}{2}- \frac{1}{6}$$
k= 2 $$\frac{1}{16-1}= \frac{1}{6}- \frac{1}{10}$$
k= 3 $$\frac{1}{36-1}= \frac{1}{10}- \frac{1}{14}$$
k= 4 $$\frac{1}{63}= \frac{1}{14}- \frac{1}{18}$$
Do you see what happens when you add those?

 

[Natasha1]

$$\Sigma_{k=1}^n\frac{1}{4k^2-1}= \frac{n}{2n+1}$$
Is that what you mean? Even if you can't use latex, please use parentheses! Most people would read 1/4k^2-1 as (1/4)k^2- 1 or 1/(4k^2)- 1 rather than 1/(4k^2-1).
Yes, your partial fraction decomposition is correct. To see how it applies to the sum, write out a few of the terms using that decomposition:
doing, say k= 1 to 4:
k= 1 $$\frac{1}{4-1}= \frac{1}{2}- \frac{1}{6}$$
k= 2 $$\frac{1}{16-1}= \frac{1}{6}- \frac{1}{10}$$
k= 3 $$\frac{1}{36-1}= \frac{1}{10}- \frac{1}{14}$$
k= 4 $$\frac{1}{63}= \frac{1}{14}- \frac{1}{18}$$
Do you see what happens when you add those?

Yes thanks I have now completed the exercise :-). Thanks!

 

[Natasha1]

$$\Sigma_{k=1}^n\frac{1}{4k^2-1}= \frac{n}{2n+1}$$
Is that what you mean? Even if you can't use latex, please use parentheses! Most people would read 1/4k^2-1 as (1/4)k^2- 1 or 1/(4k^2)- 1 rather than 1/(4k^2-1).
Yes, your partial fraction decomposition is correct. To see how it applies to the sum, write out a few of the terms using that decomposition:
doing, say k= 1 to 4:
k= 1 $$\frac{1}{4-1}= \frac{1}{2}- \frac{1}{6}$$
k= 2 $$\frac{1}{16-1}= \frac{1}{6}- \frac{1}{10}$$
k= 3 $$\frac{1}{36-1}= \frac{1}{10}- \frac{1}{14}$$
k= 4 $$\frac{1}{63}= \frac{1}{14}- \frac{1}{18}$$
Do you see what happens when you add those?

Exercise done thanks!