A sequence Problem

$a_{n+2}a_{n+1} - a_{n}a_{n+1} = 2$

$\displaystyle \sum_{n=1}^{99} a_{n+2}a_{n+1} - a_{n}a_{n+1} = \sum_{n=1}^{99} 2$

$a_{3}a_{2} - a_{2}a_{1} + a_{4}a_{3} - a_{3}a_{2} + a_{5}a_{4} - a_{4}a_{3} + \cdots a_{101}a_{100} - a_{99}a_{100} = 2\times99$

$a_{101}a_{100} - a_{2}a_{1} = 2\times99$

$a_{101}a_{100} - 2 = 2\times99$

$\boxed{a_{101}a_{100} = 2\times(99 +1) = 200}$

It is given that

$\begin{aligned} a_{n+2} & = \frac 2{a_{n+1}} + a_n & \small \color{#3D99F6} \text{Multiply both sides by }a_{n+1} \\ a_{n+2}a_{n+1} & = a_{n+1}a_n + 2 & \small \color{#3D99F6} \text{Let }a_{n+1}a_n = b_n \\ \implies b_{n+1} & = b_n + 2 & \small \color{#3D99F6} \text{Summing both sides from }k=1 \text{ to }n \\ \sum_{k=1}^n b_{k+1} & = \sum_{k=1}^n b_k + \sum_{k=1}^n 2 \\ b_{n+1} & = b_1 + 2n \\ a_{n+2}a_{n+1} & = a_2a_1 + 2n \\ & = 2+2n \\ & = 2(n+1) \\ \implies a_{n+1}a_n & = 2n \\ a_{101}a_{100} & = \boxed{200} \end{aligned}$

Let $b_{n}=a_{n}\cdot a_{n+1}$ , then, by multiplying the equation which satisfies $a_{n}$ by $a_{n+1}$ and replacing $b_{n}$ , we have: $b_{n+1}=2+b_{n} \Rightarrow b_{n}=(n-1)\cdot 2 +b_{1}$ As we know $b_{1}=a_{1}\cdot a_{2}=1\cdot 2=2$ , so we can find, by making $n=100$ : $a_{100}\cdot a_{101}=b_{100}=99\cdot 2+2=100\cdot 2=200$

$a_{n+2}a_{n+1}= 2 + a_{n+1}a_n$ and set $n=99$

Then continue the pattern and get,

$a_{100}a_{101}= 2\times 99 +a_1a_2 = 100 \times 2 = \boxed{200}$ .

After several calculations, it became apparent that a 2n x a 2n+ 1 = 4n, so a 100 x a 101 = 200. (n =50). Ed Gray