Question

`a_(1) = 1 \\ a_(n + 1) = \sqrt[]{1 + a_(n)}`

Prove by mathematical induction that the sequence is increasing​

Answer 1

Trivially, we have

`a_2 = √(1+a_1) = \sqrt2 > 1 = a_1`

so the base case is satisfied.

Assume
`a_k > a_(k-1)`, so

`a_k = \sqrt{1+a_(k-1)} > a_(k-1)`

We want to use this to show
`a_(k+1) > a_k`.
`\sqrt x` is an increasing function, so from our assumption it follows that

`a_(k+1) = √(1 + a_k) > \sqrt{1 + a_(k-1)} = a_k`

QED

Answer 2

Answer + Step-by-step explanation:

`a_(1)=1`

`a_(2)=\sqrt{1+a_(1)} =√(1+1) = √(2)`

then

`a_(2) \geq a_(1)`

`\text{suppose}\ a_(n+1) \geq a_(n)\ \text{and prove that} \ a_(n+2)\geq a_(n+1)`

`a_(n+2)=\sqrt{1+a_(n+1)} \geq \sqrt{1+a_(n)} = a_(n+1)`

Then

`a_(n+2) \geq a_(n+1)`

Then ,according to the principal of mathematical induction:

the sequence (an) is increasing​.