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[tex]a_{1} = 1 \\ a_{n + 1} = \sqrt[]{1 + a_{n}} [/tex]
Prove by mathematical induction that the sequence is increasing​

Respuesta :

LammettHash

Trivially, we have

[tex]a_2 = \sqrt{1+a_1} = \sqrt2 > 1 = a_1[/tex]

so the base case is satisfied.

Assume [tex]a_k > a_{k-1}[/tex], so

[tex]a_k = \sqrt{1+a_{k-1}} > a_{k-1}[/tex]

We want to use this to show [tex]a_{k+1} > a_k[/tex]. [tex]\sqrt x[/tex] is an increasing function, so from our assumption it follows that

[tex]a_{k+1} = \sqrt{1 + a_k} > \sqrt{1 + a_{k-1}} = a_k[/tex]

QED

Answer + Step-by-step explanation:

[tex]a_{1}=1[/tex]

[tex]a_{2}=\sqrt{1+a_{1}} =\sqrt{1+1} = \sqrt{2}[/tex]

then

[tex]a_{2} \geq a_{1}[/tex]

[tex]\text{suppose}\ a_{n+1} \geq a_{n}\ \text{and prove that} \ a_{n+2}\geq a_{n+1}[/tex]

[tex]a_{n+2}=\sqrt{1+a_{n+1}} \geq \sqrt{1+a_{n}} = a_{n+1}[/tex]

Then

[tex]a_{n+2} \geq a_{n+1}[/tex]

Then ,according to the principal of mathematical induction:

the sequence (an) is increasing​.

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