Inductions

**Alonski** · April 23rd 2008, 06:30 AM

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.

**flyingsquirrel** · April 23rd 2008, 07:45 AM

Hi

Denote $ 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} $
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} +\frac{(-1)^{\frac{N+2}{2}-1} 2(N+2)}{(N+2)^2-1} $

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 :
$S_{N+2}= \frac{N+1+(-1)^{\frac{N}{2}+1}}{N+1} +\frac{(-1)^{\frac{N+2}{2}-1} 2(N+2)}{(N+3)(N+1)}$

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

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) +(-1)^{\frac{N}{2}} 2(N+2) \right) $

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) \right) $
$S_{N+2} = \frac{1}{(N+1)(N+3)} \left(N^2+4N+3+(-1)^{\frac{N}{2}}(N+1) \right) $

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) \right) $

Hence $\boxed{S_{N+2} = \frac{N+3+(-1)^{\frac{N}{2}}}{N+3}}$

**Alonski** · April 23rd 2008, 08:57 AM

Thanks!
It took me a while to understand everything but now I get it. Awesome =)

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