Method of difference and recurrence relations

• Sep 7th 2010, 12:35 AM
helloying
Method of difference and recurrence relations
I got some same type of qns that i don't know how to solve.

Find:$\displaystyle \sum^{n}_{r=1} \frac{1}{(2n+3)(2n+1)}$
I found and checked the ans to be $\displaystyle \frac{1}{6} - \frac{1}{4n+1}$
Hence solve $\displaystyle \sum^{n}_{r=1} \frac{1}{4n^2 -1}$

And another similar one,find $\displaystyle \sum^{n}_{r=3} \frac{4-r}{r(r-1)(r-2)}$
hence find $\displaystyle \sum^{n}_{r=1} \frac{2-r}{r(r+1)(r+2)}$

I dont know how to do the hence part. Btw sorry about the summation. i cant figure out how to use latex properly

Thank you for ur time
• Sep 7th 2010, 12:30 PM
red_dog
$\displaystyle S_1=\displaystyle\sum_{r=1}^n\frac{1}{(2r+1)(2r+3) }=\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\ldots+\dfrac{1}{(2n-1)(2n+1)}+\dfrac{1}{(2n+1)(2n+3)}$

$\displaystyle S_2=\displaystyle\sum_{r=1}^n\dfrac{1}{4r^2-1}=\displaystyle\sum_{r=1}^n\dfrac{1}{(2r-1)(2r+1)}=$

$\displaystyle =\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\ldots+\dfrac{1}{(2n-1)(2n+1)}=$

$\displaystyle =\dfrac{1}{1\cdot 3}+S_1-\dfrac{1}{(2n+1)(2n+3)}$

Now replace $\displaystyle S_1$.

You can continue in the same way for the rest of the problem.