$ a_1 = 10 $ and $ a_2 =20$. Given that, $ a_{n+1} = a_{n-1} - (4/(a_n)) $ . Find smallest k for which $a_k= 0 $

Question

$ a_1 = 10 $ and $ a_2 =20$. Given that, $ a_{n+1} = a_{n-1} - (4/(a_n)) $ . Find smallest k for which $a_k= 0 $

A. Does not exist. B. is 200 C. is 50. D. is 52.

Answer 1

$$ a_{n+1} = a_{n-1} - \frac{4}{a_n}$$ $$\implies \large 4=a_{n-1}a_n - a_na_{n+1} $$ So we can write by a telescoping series; $$\begin{align} & 4=a_{n-1}a_n - a_na_{n+1} \\ & 4=a_{n-2}a_{n-1} - a_{n-1}a_{n} \\ &4=a_{n-3}a_{n-2} - a_{n-2}a_{n-1} \\ & \ldots \\ & 4=a_1a_2 - a_2a_3 \end{align}$$

By adding the $(n-1)$ sums, we get by cancelling, $$4(n-1)=a_1a_2-a_na_{n+1}$$ $$\implies \large a_na_{n+1} = -4n+204 \tag1$$

Now, observe that the product $a_na_{n+1}$ first becomes zero for $n=51$. But for $n=50$ using $(1)$, $a_{50}a_{51}=4\not=0$. And since such a product can only be $0$ if either of the terms is $0$, so $a_{51} \not=0$; rather $a_{52}=0$.

But then using $(1)$, $a_{52}a_{53}=-4\not=0 \Rightarrow a_{52}\not=0$ , neither of them, that is.

So this is a contradiction.

Hence the smallest $k$ for which $a_k=0$ does not exist i.e no such $k$ exists for $k \in \mathbb{I^+}$.