# Further Mathematics Core Pure - Series (Year 2) HELP

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#1
Will attach an image of the question; I could do part (a) and found that the partial fractions were 1/(r+2) - 2/(r+3) + 1/(r+4)
However, have no idea how to do part (b) as I don't know how to apply the method of differences to this situation. Any help will be greatly appreciated!
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#2
The question:
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1 month ago
#3
(Original post by confuzzledteen)
Will attach an image of the question; I could do part (a) and found that the partial fractions were 1/(r+2) - 2/(r+3) + 1/(r+4)
However, have no idea how to do part (b) as I don't know how to apply the method of differences to this situation. Any help will be greatly appreciated!
Write out the first few (partial fraction) terms when r=1,2,3,4 and last few terms. See which cancel.
Last edited by mqb2766; 1 month ago
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#4
(Original post by mqb2766)
Write out the first few (partial fraction) terms when r=1,2,3,4 and last few terms. See which cancel.
I tried to,, but for
r = 1: 1/3 - 1/2 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 1/3 + 1/7
r = 4: 1/6 - 2/7 + 1/8
Unlike the other questions I've done (where terms cancel in a regular pattern), I just can't seem to spot a pattern for this question..
Annoying bc I don't have the markscheme ++ in my textbook, there are never 3 terms it's always 2 haha so don't really have anything to reference to
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1 month ago
#5
(Original post by confuzzledteen)
I tried to,, but for
r = 1: 1/3 - 1/2 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 1/3 + 1/7
r = 4: 1/6 - 2/7 + 1/8
Unlike the other questions I've done (where terms cancel in a regular pattern), I just can't seem to spot a pattern for this question..
Annoying bc I don't have the markscheme ++ in my textbook, there are never 3 terms it's always 2 haha so don't really have anything to reference to
Just look down the diagonals
1/5 - 2/5 + 1/5 = ?
...
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1 month ago
#6
(Original post by confuzzledteen)
I tried to,, but for
r = 1: 1/3 - 1/2 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 1/3 + 1/7
r = 4: 1/6 - 2/7 + 1/8
Unlike the other questions I've done (where terms cancel in a regular pattern), I just can't seem to spot a pattern for this question..
Annoying bc I don't have the markscheme ++ in my textbook, there are never 3 terms it's always 2 haha so don't really have anything to reference to
When trying to spot a pattern, it's a good idea NOT to cancel (simplify) the fractions - the simpifications occur somewhat "randomly" and occur the pattern.

If you replace the 1/2 and 1/3 terms in the center column by 2/4 and 3/6 it may make it easier to spot the pattern here.
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#7
(Original post by DFranklin)
When trying to spot a pattern, it's a good idea NOT to cancel (simplify) the fractions - the simpifications occur somewhat "randomly" and occur the pattern.

If you replace the 1/2 and 1/3 terms in the center column by 2/4 and 3/6 it may make it easier to spot the pattern here.
Oh cool, thanks! I see the pattern now.. Thanks both of you guys 0
#8
Oh wait sorry, when I cross out the common terms, I end up with
r = 1: 1/3 - 2/4 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 2/6 + 1/7
r = 4: 1/6 - 2/7 + 1/8
r = 5: 1/7 - 2/8 + 1/9
r = n: 1/(n+2) - 2/(n+3) + 1/(n+4)
(bold terms representing those that have been crossed out)
I'm unsure of where to go from here though, as usually we only have one term left + some fraction involving n but in this case, we've got many?
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1 month ago
#9
(Original post by confuzzledteen)
Oh wait sorry, when I cross out the common terms, I end up with
r = 1: 1/3 - 2/4 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 2/6 + 1/7
r = 4: 1/6 - 2/7 + 1/8
r = 5: 1/7 - 2/8 + 1/9
r = n: 1/(n+2) - 2/(n+3) + 1/(n+4)
(bold terms representing those that have been crossed out)
I'm unsure of where to go from here though, as usually we only have one term left + some fraction involving n but in this case, we've got many?
Cross out the full diagonal. The three numerator are
+1 -2 +1
Whatever the denominator.

You cant cross out two positive terms. One (at least) must be negative.
Last edited by mqb2766; 1 month ago
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#10
(Original post by mqb2766)
Cross out the full diagonal. The three numerator are
+1 -2 +1
Whatever the denominator.
Ohhhh, alright - thank you so much! So would that leave me with:

r = 1: 1/3 - 2/4 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 2/6 + 1/7
r = 4: 1/6 - 2/7 + 1/8
r = 5: 1/7 - 2/8 + 1/9
r = n: 1/(n+2) - 2/(n+3) + 1/(n+4)

so would the final answer just be (after simplifying): 1/3 - 2/4 + 1/6 - 2/(n+3) + 1/(n+4) ?
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1 month ago
#11
(Original post by confuzzledteen)
Ohhhh, alright - thank you so much! So would that leave me with:

r = 1: 1/3 - 2/4 + 1/5
r = 2: 1/4 - 2/5 + 1/6
r = 3: 1/5 - 2/6 + 1/7
r = 4: 1/6 - 2/7 + 1/8
r = 5: 1/7 - 2/8 + 1/9
r = n: 1/(n+2) - 2/(n+3) + 1/(n+4)

so would the final answer just be (after simplifying): 1/3 - 2/4 + 1/6 - 2/(n+3) + 1/(n+4) ?
Why not just check carefully the final expression? Then repost.
Note you should write out the last few lines as well as the first few.
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#12
(Original post by mqb2766)
Why not just check carefully the final expression? Then repost.
Note you should write out the last few lines as well as the first few.
And my final answer was n(n-5)/12(n+3)(n+4), so a=1, b=-5 and c=12
I don't have the markscheme so I can't check if it's right or not,, but I feel like it is as the constant terms cancelled out
Thanks so much for all ur help x
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1 month ago
#13
(Original post by confuzzledteen)
And my final answer was n(n-5)/12(n+3)(n+4), so a=1, b=-5 and c=12
I don't have the markscheme so I can't check if it's right or not,, but I feel like it is as the constant terms cancelled out
Thanks so much for all ur help x
Can you post your working for the last part?
One way to check is to put n as a low number. n=1 is the lowest/easiest.
Last edited by mqb2766; 1 month ago
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#14
(Original post by mqb2766)
Can you post your working for the last part?
One way to check is to put n as a low number. n=1 is the lowest/easiest.
Sure:

(1/3) - (2/4) + (1/4) - (2/n+3) + (1/n+4)
= (1/12) - (2/n+3) + (1/n+4)
= [ (n+3)(n+4) - 2(12)(n+4) + 1(12)(n+3) ] / 12(n+3)(n+4)
= [ (n^2 + 7n + 12) - (24n + 48) + (12n + 36) ] / 12(n+3)(n+4)
= n^2 - 5n / 12(n+3)(n+4)
= n(n-5) / 12(n+3)(n+4)

When I put n as a low number, in to my solution, I get -1/60
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1 month ago
#15
(Original post by confuzzledteen)
Sure:

(1/3) - (2/4) + (1/4) - (2/n+3) + (1/n+4)
= (1/12) - (2/n+3) + (1/n+4)
= [ (n+3)(n+4) - 2(12)(n+4) + 1(12)(n+3) ] / 12(n+3)(n+4)
= [ (n^2 + 7n + 12) - (24n + 48) + (12n + 36) ] / 12(n+3)(n+4)
= n^2 - 5n / 12(n+3)(n+4)
= n(n-5) / 12(n+3)(n+4)

When I put n as a low number, in to my solution, I get -1/60
Err, that -1/60 must concern you? You're adding up positive numbers and get a negative value?
N=1 must correspond to the first value 1/30.

You didn't include the terms in the last few lines correctly, as mentioned previously. You've only looked at the last line, not n-1,...
Last edited by mqb2766; 1 month ago
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#16
(Original post by mqb2766)
Err, that -1/60 must concern you? You're adding up positive numbers and get a negative value?
N=1 must correspond to the first value 1/30.

You didn't include the terms in the last few lines correctly, as mentioned previously. You've only looked at the last line, not n-1,...
Oh damn, I didn't even notice.. Thank you so much. I just did it again and got n(n+7)/12(n+3)(n+4) which gives me 1/30 when I sub n=1 in. Never knew about this trick, so thank you loads.

I've got a question though; for the question we just did, why did I have to consider r = n-1 as well? As for some questions (like the one I just attached), I only had to consider r =1, r = 2, r = 3 and then r = n. Why is there a need to consider r = n-1 for some questions (and not for others, like the one I attached) ?
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1 month ago
#17
(Original post by confuzzledteen)
Oh damn, I didn't even notice.. Thank you so much. I just did it again and got n(n+7)/12(n+3)(n+4) which gives me 1/30 when I sub n=1 in. Never knew about this trick, so thank you loads.

I've got a question though; for the question we just did, why did I have to consider r = n-1 as well? As for some questions (like the one I just attached), I only had to consider r =1, r = 2, r = 3 and then r = n. Why is there a need to consider r = n-1 for some questions (and not for others, like the one I attached) ?
That sounds right.

There isn't a magic "you must include xxx lines". In this case you were cancelling out terms (summing to zero) along a three diagonal. It's fairly easy to see that terms from the first two and last two lines will not cancel. Think logically and there is generally some symmetry to the problem.

What did you do for the extra question? Can you just write out the first few and last few lines? I'm guessing only the first and last lines have nonzero terms after cancelling? Note the denominator is a single term.
Last edited by mqb2766; 1 month ago
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#18
(Original post by mqb2766)
That sounds right.

There isn't a magic "you must include xxx lines". In this case you were cancelling out terms (summing to zero) along a three diagonal. It's fairly easy to see that terms from the first two and last two lines will not cancel. Think logically and there is generally some symmetry to the problem.

What did you do for the extra question? Can you just write out the first few and last few lines?
Ahh alright, thanks! For the extra question, I first split 3/r(r+1) into partial fractions, which are 3/r and 3/(r+1).. I then did
r = 1 : 3 - 3/2
r = 2 : 3/2 - 3/3
r = 3 : 3/3 - 3/4
r = n : 3/n - 3/n+1
Bold terms cancel out..
So i ended up with 3 - 3/n+1 which I simplified: [ 3(n+1) - 3 ] / n+1 = 3n/n+1
Using the technique you taught me about, subbed in n=1 and got 3/2 which I should 0
1 month ago
#19
(Original post by confuzzledteen)
Ahh alright, thanks! For the extra question, I first split 3/r(r+1) into partial fractions, which are 3/r and 3/(r+1).. I then did
r = 1 : 3 - 3/2
r = 2 : 3/2 - 3/3
r = 3 : 3/3 - 3/4
r = n : 3/n - 3/n+1
Bold terms cancel out..
So i ended up with 3 - 3/n+1 which I simplified: [ 3(n+1) - 3 ] / n+1 = 3n/n+1
Using the technique you taught me about, subbed in n=1 and got 3/2 which I should You must see here that the "triangular blocks" of non cancelled elements at the start and end each just contain a single element. Hence writing down only line n is correct, but risky if you can't easily see the pattern. Generally if you had non cancelled elements in the first two rows, the same would occur on the last two. Similarly for first 3, last 3,... You're cancelling a large number of diagonals which takes a large parallelogram of entries out, leaving two triangles, at the top and bottom, of non cancelled entries.
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#20
(Original post by mqb2766)
You must see here that the "triangular blocks" of non cancelled elements at the start and end each just contain a single element. Hence writing down only line n is correct, but risky if you can't easily see the pattern. Generally if you had non cancelled elements in the first two rows, the same would occur on the last two. Similarly for first 3, last 3,... You're cancelling a large number of diagonals which takes a large parallelogram of entries out, leaving two triangles, at the top and bottom, of non cancelled entries.
Oh, that's super smart. That makes sense.. I had always just assumed it was a random guess or something lol - the explanation behind it is actually super interesting. Again, thanks so much for all your help!
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