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Suppose $$n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}$$ for every postive integers $n\ge 1$,Give that $a_{0}=1,a_{1}=2$

find the $a_{n}$

My try:

$$a_{2}=\dfrac{1}{2}=\dfrac{1}{2!},a_{3}=\dfrac{1}{6}=\dfrac{1}{3!},\cdots$$ so I guess $$a_{n}=\dfrac{1}{n!}(n\ge 2)$$ and we easy use the Mathematical induction prove it.

My question:

This problem have without mathematical induction to solve it?

and following is my idea:

let $n(n+1)a_{n+1}=b_{n+1}$ so $$b_{n+1}=b_{n}-\dfrac{b_{n-1}}{n-1},n\ge 2$$ then I can't,Thank you for your help!

This problem from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1290603&sid=5c95d4057706d4a61097b007952afb70#p1290603

math110
  • 93,304

1 Answers1

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Let $c_n = na_n -a_{n-1}$. Then, rearranging:

$$n[(n+1)a_{n+1}-(n-1)a_n] = -(n-2)a_{n-1}$$

$$n[(n+1)a_{n+1}-a_n - (n-2)a_n] = -(n-2)a_{n-1}$$

$$nc_{n+1} - n(n-2)a_n = -(n-2)a_{n-1}$$

$$nc_{n+1} = (n-2)(na_n - a_{n-1})$$

$$c_{n+1} = (1-2/n)\cdot c_n$$

Plugging in $n=2$, we have $c_3=0$, and it follows that $c_4=0$, $c_5=0$, etc. So, for $n\geq 3$, $c_n=0$ and $na_n = a_{n-1}$.

We have to calculate one value "by hand": plugging in $n=1$ into the original equation gives $a_2=1/2$.

Then, for $n\geq 2$, we calculate: $a_n = 1/n \cdot a_{n-1} = 1/n \cdot 1/(n-1) \cdot \ldots \cdot a_2 = 1/n!$.

Andrew Dudzik
  • 30,074
  • +1, especially for taking care of the inconvenience that the prefactor $1-2/n$ is zero for some $n$. – Did Nov 03 '13 at 12:12
  • @Did Hmm, it seems more like a convenience that I didn't take advantage of. I confused the problem with the work below, and assumed that $a_2=1/2$, $a_3=1/3$ were initial conditions. In fact, you only need $a_2$, and technically, you have to deduce it from the problem somehow. Fixed. – Andrew Dudzik Nov 03 '13 at 20:34