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Math Help - Inductions

  1. #1
    Newbie
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    Apr 2008
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    Inductions

    I have this problem:
    4/3-8/15+12/35-16/63+...+((-1)^(1/2N-1)*2N)/(N^2-1)=(N+1+(-1)^(1/2N+1))/(N+1)

    Works for every whole n that is an even number.

    I checked if it works when n=2 and it does.

    I said that when n=k it works.

    While proving that it works when n=k+2 I got it down all the way to this but can't figure out how to proceed:
    (K+1+(-1)^(1/2K+1))/(K+1)-((1)^(1/2K-1/2)*(2K+2))/(K^2+4K+4)=(K+3+(-1)^(1/2K+2))/(K+3)

    I also included an image so that it will be easier to read.

    Please help and if you can include explanations of every step that would be great.
    Attached Thumbnails Attached Thumbnails Inductions-mathproblem2.bmp  
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
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    802
    Hi

    Denote <br />
  S_{N+2}=\frac{4}{3}-\frac{8}{15}+\frac{12}{35}-\frac{16}{63}+\ldots+ \frac{(-1)^{\frac{N}{2}-1}\times2N}{N^2-1}  +\frac{(-1)^{\frac{N+2}{2}-1}\times 2(N+2)}{(N+2)^2-1}<br />
    As S_N=\frac{N+1+(-1)^{\frac{N}{2}+1}}{N+1}, the sum becomes S_{N+2}= \frac{N+1+(-1)^{\frac{N}{2}+1}}{N+1}<br />
+\frac{(-1)^{\frac{N+2}{2}-1} 2(N+2)}{(N+2)^2-1}<br />

    First, let's factor the denominator of the second fraction : (N+2)^2-1=(N+2-1)(N+2+1)=(N+1)(N+3) hence :
    S_{N+2}= \frac{N+1+(-1)^{\frac{N}{2}+1}}{N+1}<br />
+\frac{(-1)^{\frac{N+2}{2}-1} 2(N+2)}{(N+3)(N+1)}

    Factor by \frac{1}{N+1} :
    S_{N+2} = \frac{1}{N+1} \left( N+1+(-1)^{\frac{N}{2}+1}<br />
+\frac{(-1)^{\frac{N+2}{2}-1} 2(N+2)}{N+3} \right)<br />
    S_{N+2} = \frac{1}{N+1} \left( \frac{(N+3)\left(N+1+(-1)^{\frac{N}{2}+1}\right)}{N+3}<br />
+\frac{(-1)^{\frac{N}{2}} 2(N+2)}{N+3} \right)<br />

    Factor by \frac{1}{N+3} and develop the product :
    S_{N+2} = \frac{1}{(N+1)(N+3)} \left(N^2+N+3N+3+(-1)^{\frac{N}{2}+1}(N+3)<br />
+(-1)^{\frac{N}{2}} 2(N+2) \right)<br />

    Regroup the terms which have to do with (-1)^{\frac{N}{2}}, acknowledged that (-1)^{\frac{N}{2}+1}=-(-1)^{\frac{N}{2}} :
    S_{N+2} = \frac{1}{(N+1)(N+3)} \left(N^2+4N+3+(-1)^{\frac{N}{2}}\left(2(N+2)-N-3\right)<br />
 \right)<br />
    S_{N+2} = \frac{1}{(N+1)(N+3)} \left(N^2+4N+3+(-1)^{\frac{N}{2}}(N+1)<br />
 \right)<br />

    Here, notice that (-1)^2+4\times(-1)+3=0 : -1 is a root of the polynomial whence N^2+4N+3=(N+1)(aN+b) Developping the expression and identifying the coefficients yields b=3 and a=1\,\Rightarrow N^2+4N+3=(N+1)(N+3). Replace this in the expression of S_{N+2} :
    S_{N+2} = \frac{1}{(N+1)(N+3)} \left((N+1)(N+3)+(-1)^{\frac{N}{2}}(N+1)<br />
 \right)<br />

    Hence \boxed{S_{N+2} = \frac{N+3+(-1)^{\frac{N}{2}}}{N+3}}
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  3. #3
    Newbie
    Joined
    Apr 2008
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    Thanks!
    It took me a while to understand everything but now I get it. Awesome =)
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