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Step II 1995 Q.1 watch

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     1-2+3-4+...+(-1)^{n-1}n = - \dfrac{n}{2} when n is even.

     1-2+3-4+...+(-1)^{n-1}n =  \dfrac{n+1}{2} when n is odd.

    iii) Show that  1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2 = (-1)^{n-1}(An^2 + Bn) where the constants A and B are to be determined.

    Not really sure how to approach it. Any help is much appreciated.
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    (Original post by Clarity Incognito)
    ...
    There are solutions referenced from the BIG FAT STEP MEGA THREAD in Maths Exams.

    Unfortunately not all the links seem to point to the right place.

    BIG SPOILER ALERT:

    See post #18 in http://www.thestudentroom.co.uk/showthread.php?t=359567
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    (Original post by ghostwalker)
    There are solutions referenced from the BIG FAT STEP MEGA THREAD in Maths Exams.

    Unfortunately not all the links seem to point to the right place.

    BIG SPOILER ALERT:

    See post #18 in http://www.thestudentroom.co.uk/showthread.php?t=359567
    Yeah, I found that some of the links led to different solutions too. I didn't want to look at the answer straight away because I wanted to play around with one little hint before consulting the solutions. If you could offer a hint, that would be terribly kind!
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    (Original post by Clarity Incognito)
    Yeah, I found that some of the links led to different solutions too. I didn't want to look at the answer straight away because I wanted to play around with one little hint before consulting the solutions. If you could offer a hint, that would be terribly kind!
    Not done the question, so .... But you might want to post the first part of the question, then someone else might ....
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    (Original post by ghostwalker)
    Not done the question, so .... But you might want to post the first part of the question, then someone else might ....
    I'm pretty sure that the above information is all that is required, the rest is irrelevant but I'm tired so I'm just going to cheat now :o: .
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    I haven't seen the question either so I'm not sure if this approach is OK.

    Look at the first few sums

    1, -3, 6, -10, 15.....

    and prove the obvious formula by induction.
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    (Original post by rnd)
    I haven't seen the question either so I'm not sure if this approach is OK.

    Look at the first few sums

    1, -3, 6, -10, 15.....

    and prove the obvious formula by induction.
    I'm interested to see whether you think you can extrapolate your quoted method to the last part in the question.

    The background of the question is  1 +x + x^2+...+x^n = \dfrac{1-x^{n+1}}{1-x} Differentiating both sides and putting x = -1. The first two sums in my first post follow shortly.

    It's this change with each term being squared on the LHS, I wasn't sure how to apply this change to the RHS.

    Anyway, I'd appreciate any insight you could provide.
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    (Original post by Clarity Incognito)
    I'm interested to see whether you think you can extrapolate your quoted method to the last part in the question.

    The background of the question is  1 +x + x^2+...+x^n = \dfrac{1-x^{n+1}}{1-x} Differentiating both sides and putting x = -1. The first two sums in my first post follow shortly.

    It's this change with each term being squared on the LHS, I wasn't sure how to apply this change to the RHS.

    Anyway, I'd appreciate any insight you could provide.
    Hah, by coincidence I'm on this question too, and a little stuck :P

    What I'm about to do is try to multiply the derivative we found by x, and then differentiate again - that gives the coefficients of all the squares, and let x = -1 again. Not sure if it'll work but I'll report back
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    I think the natural approach is to think how you managed to get the constants in front, and maybe repeat this same process?

    d(x^n)/dx=n(n-1)x^n-2
    you want to get the n(n-1) as n^2.

    If you wrote out the line in x (before letting x=-1 for the previous part) then the working follows from that.
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    Have to admire people who get S or even 1 in STEP II, to be fair.
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    (Original post by ziedj)
    Hah, by coincidence I'm on this question too, and a little stuck :P

    What I'm about to do is try to multiply the derivative we found by x, and then differentiate again - that gives the coefficients of all the squares, and let x = -1 again. Not sure if it'll work but I'll report back
    Yeah, thats the approach.
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    (Original post by ziedj)
    Hah, by coincidence I'm on this question too, and a little stuck :P

    What I'm about to do is try to multiply the derivative we found by x, and then differentiate again - that gives the coefficients of all the squares, and let x = -1 again. Not sure if it'll work but I'll report back


    Surely, we'd want x = 1 to conserve the signs though?

    Good thinking btw.
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    (Original post by Clarity Incognito)


    Surely, we'd want x = 1 to conserve the signs though?

    Good thinking btw.
    The signs are alternating. I've done the differentiation, and using x = -1 gets exactly what I want on the left - but oh my days, the right hand side is just a MESS I can hardly keep track of what my -1s are doing haha. But yeah, using x = -1 gets exactly what you need on the LHS.
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    (Original post by Clarity Incognito)
    I'm interested to see whether you think you can extrapolate your quoted method to the last part in the question.
    OK I found the paper now.

    http://www.mathshelper.co.uk/STEP%20II%201995.pdf

    What I wrote was for the last part of the question.

    It certainly works but is it acceptable? I don't know.
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    (Original post by ziedj)
    The signs are alternating. I've done the differentiation, and using x = -1 gets exactly what I want on the left - but oh my days, the right hand side is just a MESS I can hardly keep track of what my -1s are doing haha. But yeah, using x = -1 gets exactly what you need on the LHS.
    Yeah, I got confused with what I was multiplying by x and what needed to be differentiated. I'll post my RHS after I've done it.

    (Original post by rnd)
    OK I found the paper now.

    http://www.mathshelper.co.uk/STEP%20II%201995.pdf

    What I wrote was for the last part of the question.

    It certainly works but is it acceptable? I don't know.
    Sweet. Will try it this way nonetheless.
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    At first glance if you factorise as difference of 2 squares you get something like
    (1-2)(1+2) + (3-4)(3+4) + (5-6)(5+6)+...
    =-1(1+2+3+4+5+6+...

    Given that we know the term in brackets I think this should work.

    But again I haven't tried it properly, and far too tired to put pen to paper now.

    EDIT: Couldn't sleep with this on my mind, but it does indeed work. You have to consider n odd and n even obviously, but the brackets form an AP. Might want to write 1+2 as 3 instead though to make it obvious that we have -1(3+7+11+...

    Edit2:
    Spoiler:
    Show
    Looking back at my solution I was "wrong" in the previous edit... It's easier to write
    -1(1+2+3+...+n) whereas in my previous edit I was suggesting to write -1(3+7+...+(2n-1))
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    Note that you *might* lose marks for assuming the sum of an AP (since the question asks you to *derive* the sum of GP in the first part, rather than just assuming it).
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    (Original post by rnd)
    OK I found the paper now.

    http://www.mathshelper.co.uk/STEP%20II%201995.pdf

    What I wrote was for the last part of the question.

    It certainly works but is it acceptable? I don't know.
    was just looking at speleo's solutions from this thread (post 18):http://www.thestudentroom.co.uk/showthread.php?t=359567 for part ii, i dont understand how he/she deduced the following:

    Spoiler:
    Show
    If n is odd,
    S = 2(n+1)/4 = (n+1)/2
    If n is even,
    S = [-2(n+1) - 2]/4
    = [-2n - 2 + 2]/4 = n/2


    the first three lines are straight forward, it's just differentiating LHS and RHS and substituting x=-1 but beyond that, no clue. just test odd/even numbers into the differentiated equation?
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    (Original post by KeineHeldenMehr)
    was just looking at speleo's solutions from this thread (post 18):http://www.thestudentroom.co.uk/showthread.php?t=359567 for part ii, i dont understand how he/she deduced the following:

    Spoiler:
    Show
    If n is odd,
    S = 2(n+1)/4 = (n+1)/2
    If n is even,
    S = [-2(n+1) - 2]/4
    = [-2n - 2 + 2]/4 = n/2


    the first three lines are straight forward, it's just differentiating LHS and RHS and substituting x=-1 but beyond that, no clue. just test odd/even numbers into the differentiated equation?
    Essentially, dependent on whether n is odd or even, the following sums are deduced as all the -1's in the RHS that are raised to the power of (n+1) or whatever can be defined.
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    (Original post by Clarity Incognito)
    Essentially, dependent on whether n is odd or even, the following sums are deduced as all the -1's in the RHS that are raised to the power of (n+1) or whatever can be defined.
    pff...still dont get it, S = 2(n+1)/4 magically appeared.
 
 
 
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