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Method of differences

https://imgur.com/a/hnm8HO4

Where did the terms circled in red go?
Sorry I can't help you, but could you possibly tell me where you got the questions from ( ie which board, question paper etc.)?
Original post by bluenotewitt
Sorry I can't help you, but could you possibly tell me where you got the questions from ( ie which board, question paper etc.)?


Mei fp1 2016
Original post by JacobPowell
https://imgur.com/a/hnm8HO4

Where did the terms circled in red go?


It’s the method of differences... they cancel!
Original post by JacobPowell
Mei fp1 2016

Thanks!!!
Original post by RDKGames
It’s the method of differences... they cancel!


With which terms?
Original post by JacobPowell
With which terms?

Just an observation:

1/20 = 1/5- 3/20

1/28 = 1/7- 3/28

1/36 = 1/9 - 3/36
(edited 5 years ago)
Original post by JacobPowell
https://imgur.com/a/hnm8HO4

Where did the terms circled in red go?

Maybe I can help you after all... :smile:

Clearly nothing is going to cancel in the usual straight forward way, so I tried rewriting the numbers. As I observed in my previous post, one can write the terms you circled in terms of the other terms in the series, that is ( sorry, too useless to use latex !)

1/20 = 1/5- 3/20

1/28 = 1/7- 3/28

1/36 = 1/9 - 3/36...

Substituting the above into the series, gives the following:

3/4 - 1/3 + (1/5- 3/20) + 3/12 - 1/5 + (1/7- 3/28) + 3/20 - 1/7 + (1/9 - 3/36) +....

Now the terms cancel ! And we are left with the final terms in "n" as given in the markscheme and 3/4 - 1.3 + 3/12 = 3/4 -1/3 + 1/4.
Tricky question for an A-level though!
(edited 5 years ago)
Original post by bluenotewitt
Maybe I can help you after all... :smile:

Clearly nothing is going to cancel in the usual straight forward way, so I tried rewriting the numbers. As I observed in my previous post, one can write the terms you circled in terms of the other terms in the series, that is ( sorry, too useless to use latex !)

1/20 = 1/5- 3/20

1/28 = 1/7- 3/28

1/36 = 1/9 - 3/36...

Substituting the above into the series, gives the following:

3/4 - 1/3 + (1/5- 3/20) + 3/12 - 1/5 + (1/7- 3/28) + 3/20 - 1/7 + (1/9 - 3/36) +....

Now the terms cancel ! And we are left with the final terms in "n" as given in the markscheme and 3/4 - 1.3 + 3/12 = 3/4 -1/3 + 1/4.
Tricky question for an A-level though!


Ah yeah that’s the trickiest question I’ve seen on this topic so far... thanks for your help :smile:
Original post by JacobPowell
Ah yeah that’s the trickiest question I’ve seen on this topic so far... thanks for your help :smile:
The thing to do is look at the partial fraction decomposition: you have a 3/4 term, a 1/4 term and a -1 term,
(3412r1,1412r+3,112r+1\frac{3}{4}\frac{1}{2r-1}, \frac{1}{4} \frac{1}{2r+3}, -1 \frac{1}{2r+1}), which gives you a clue about how the terms are going to cancel.

It would also probably help to keep the 3/4, 1/4 fractions in when writing out (rather than multiplying 1/4 x 1/5 = 1/20 for example):

34×1113+14×15+\frac{3}{4} \times \frac{1}{1} - \frac{1}{3} + \frac{1}{4} \times \frac{1}{5} +

34×1315+14×17+\frac{3}{4}\times \frac{1}{3} - \frac{1}{5} + \frac{1}{4} \times \frac{1}{7} +

34×1517+14×19+...\frac{3}{4}\times \frac{1}{5} - \frac{1}{7} + \frac{1}{4} \times \frac{1}{9} + ...

makes it easier to spot how the 15\frac{1}{5} terms cancel etc.

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