Let `(a_n)` be a sequence of real number satisfied that: `a_0=1` and `a_(n+1)=a_0a_1......a_n+4` for all natural number n. Prove that: `a_n-\sqrt{a_{n+1}}=2` for all `n>0`

 Let `(a_n)` be a sequence of real number satisfied that: `a_0=1` and 

`a_(n+1)=a_0a_1......a_n+4` for all natural number n.

Prove that: `a_n-\sqrt{a_{n+1}}=2` for all `n>0`.

Solution

We observation that `a_k>0` for all natural number `n`

And from `a_(n+1)=a_0a_1......a_n+4`

Replace `n=n+1` then `a_(n+2)=a_0a_1......a_(n+1)+4`

                                    `a_(n+2)=(a_0a_1.....a_n)(a_0a_1....a_n+4)+4`

                                    `a_(n+2)=(a_0a_1.....a_n)^2+4(a_0a_1.....a_n)+4`

                                    `a_(n+2)=(a_0a_1....a_n+2)^2`

Therefore,     `a_(n+2)^(1/2)=a_0a_1....a_n+2`

                        `(a_(n+2))^(1/2)=(a_(n+1)+2)` or `(a_(n+1))^(1/2)=(a_n+2)`

Hence, the problem is solved.

Solution by: Thin Sokkean