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Given that $a_{n+1}=\dfrac{a_{n-1}}{1+n\cdot a_{n-1}a_{n}}$ for $n=1,2,3,....$ and $a_0=a_1=1$. Find the value of $a_{199}\cdot a_{200}$. Also give with proof the general formula of $a_{n}a_{n+1}$?

Popular Power
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3 Answers3

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You can prove the formula you found, by induction:

given $a_{n+1}=\dfrac{a_{n-1}}{1+n\cdot a_{n-1}a_{n}}$, it follows that $\dfrac1{a_{n+1}a_n}=\dfrac1{a_{n-1}a_n}+n.$

J. W. Tanner
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It is easy enough to confirm your formula.

Define $$b_n=\frac 1{a_na_{n-1}}-1$$

We have $b_1=0$ and we wish to show that $$b_n=b_{n-1}+n-1$$

You have checked that equality holds for small $n$, let's proceed by induction.

But we write out both sides to see that we are trying to show $$\frac 1{a_na_{n-1}}=\frac 1{a_{n-1}a_{n-2}}+n-1$$

Clearing denominators gives us $$a_{n-2}=a_n+a_na_{n-1}a_{n-2}(n-1)=a_n(1+a_{n-1}a_{n-2}(n-1))$$ $$\iff a_n=\frac {a_{n-2}}{1+a_{n-1}a_{n-2}(n-1)}$$

which is the defining relation for the $a_i$. And we are done.

lulu
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I did the find the general formula for $a_{n}\cdot a_{n+1}$ but I do not have a proof.

  1. $\dfrac{1}{a_0a_1}=1$
  2. $\dfrac{1}{a_1a_{2}}=2$
  3. $\dfrac{1}{a_2a_3}=4$
  4. $\dfrac{1}{a_3a_4}=7$
  5. $\dfrac{1}{a_4a_5}=11$

I think it is trivial to see the pattern.

$$\dfrac{1}{a_{n}\cdot a_{n+1}}=1+(1+2+3....n)=1+\dfrac{n(n+1)}{2}$$

But however as I earlier said, I don't have a sound proof.

Popular Power
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    While I'm not saying you found or didn't find the answer, I think it's a bit disingenuous - or at best, a leap of faith - to claim you found the general formula, when you do not prove it and evidently cannot prove it yourself. There's no inherent reason, a priori, why a pattern should hold outside of the cases you checked. – PrincessEev Mar 17 '20 at 18:18
  • I know, but it is the current answer. I wrote a program in python for checking if my pattern is correct, and it is indeed correct as far as I have checked.(above 1000) – Popular Power Mar 17 '20 at 18:20
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    @EeveeTrainer: this is an overstatement. There is sufficient evidence to support the conjecture. –  Mar 17 '20 at 18:25
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    Not sure why people are downvoting this, so I upvoted it to compensate. I guess an argument could be made that it more properly belongs within the original post, but that aside it's clear that you have made the critical observation needed to solve the problem and that you have tested it sufficiently to establish confidence, even if you fall short of a full proof. – lulu Mar 17 '20 at 18:34