Question

Formulas and Identities #1
Question 33
It is given that \(\frac{A}{k-1}-\frac{B}{k-2} = \frac{1}{(k-1)(k-2)}\), where k ≠ 1 and k ≠ 2
(ai) Show that A(k-2) - B(k-1) =1
(aii) Hence, find the values of A and B
(b) Hence, find \(\frac{1}{100 \times 102} + \frac{1}{102 \times 104} + \frac{1}{104 \times 106} + ... + \frac{1}{198 \times 200}\)
For (a), I've done that. Result of (aii) is A= -1 and B=-1
How to do (b) ?

Answer 1

i am not sure both A and B = -1, i will bet that this is a telescoping sum

Answer 2

yes, it is
\[\frac{1}{(k-1)(k-2)}=\frac{1}{k-2}-\frac{1}{k-1}\]

Answer 3

\[\frac12\left(\frac{1}{100} - \frac{1}{102} + \frac{1}{102} + \ldots +\frac{1}{200}\right)\]

Answer 4

substitute starting with \(k=101\) and see what you get. the left hand side will be the question your are given, the right hand side will telescope

Answer 5

@Aerospace.Ekaansh why the \(\frac{1}{2}\)?

Answer 6

I'm lost.
Consider the first term for part (b)
\[\frac{1}{100 \times 102}\]and use \(\large \frac{1}{(k-1)(k-2)}=\frac{1}{k-2}-\frac{1}{k-1}\) for k=101
\[ \frac{1}{(101-1)(101-2)}= \frac{1}{(100)(99)} = weird\]

Answer 7

you have some inconsistencies in your original question. I'm betting you have some typos.

Answer 8

I'm sorry... There are some mistakes in question (ai). It's now corrected

Answer 9

Ignoring all of the question and just considering the series,
\[\frac{1}{100 \times 102} + \frac{1}{102 \times 104} + \frac{1}{104 \times 106} + ... + \frac{1}{198 \times 200} = \sum \frac{1}{k(k+2)}\]\[ = \sum \frac12\left(\frac1k-\frac1{k+2}\right)\]

Answer 10

Sorry, but how did you get the last thing?
And also, I don't think we can ignore all of the question due to the word 'hence'

Answer 11

notice that all the numbers in the denominator are even...this tells us that

\[\sum \frac{1}{k(k+2)}\]

probably isn't the sum we want (unless you are letting k increase by 2's)

Answer 12

Sorry... I meant this: \(\sum \frac12\left(\frac1k-\frac1{k+2}\right)\)
And this: \( \sum \frac{1}{k(k+2)}\) should not be something I want for this question.

Answer 13

I am sorry. :/

Answer 14

\[\sum\frac {1}{(100 + 2(k-1))(100+2k)}\]?

Answer 15

that would work...we should put limits on k

Answer 16

oh wow my answer was all messed up
although the partial fractions was right

Answer 17

I don't see a direct connection from the first part of the question to the second part.

Answer 18

\[\frac{1}{2k(2k+2)}=\frac{1}{4}\frac{1}{k(k+1)} \] maybe?

Answer 19

ah

Answer 20

then use partial fractions, then start at 50 that is my guess but i messed up earlier so who knows

Answer 21

\[\frac{1}{4}\left(\frac{1}{k}-\frac{1}{k+1}\right)\]

Answer 22

Thanks I got it

Answer 23

sorry about messing up the first one
good that zarkon is here to keep us honest

Answer 24

you found the last connection :)

Answer 25

Hmm.. to be honest.. I got it because I've found my work when you're posting the solution.. Sorry for that... And thanks @Zarkon and @satellite73

Answer 26

so we are worthless ...man... ;)