# Pls demonstrate 0<xn<1 if x(n+1)=(xn^5+xn)/4, 0<x1<1? pls is urgent not try values

*print*Print*list*Cite

### 1 Answer

You need to use mathematical induction principle since the problem shows that the statement holds for the lowest term of the sequence, such that `0<x_1<1` .

You only need to perform the inductive step, hence, you need to consider that the statement holds for `x_n` and you need to prove that the statement also holds for `x_(n+1).`

`0 < x_n < 1 => 0^5 < (x_n)^5 < 1^5 => 0 < (x_n)^5 < 1 => 0 +x_n< (x_n)^5 + x_n< 1 + x_n`

But since `x_n in (0,1) => (x_n)^5 + x_n in (0,1).`

Dividing by 4 yields:

`0/4 < ((x_n)^5 + x_n)/4 < 1/4 < 1`

You need to substitute `x_(n+1)` for `((x_n)^5 + x_n)/4` , such that:

`0 < x_(n+1) < 1`

Hence, the inductive step shows that the inequality`0 < x_(n+1) < 1` holds.

**Hence, checking if `0 < x_n < 1` , using the mathematical induction principle, yields that `0 < x_n < 1` holds for `n>=1` .**