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
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