Consider the following recursively defined sequence a1 = 2 an = an-1 – 3 for n >= 2 In this question, we will prove that an = -3n + 5 for all n >=1. Fill the blanks

Let P(n) = ______. We prove that for all n there exists Z, if n>=0

Base case: The base case is n = _

Proof of the base case: ______

Induction step: The induction hypothesis is _ _

Our goal in the induction is to prove

Proof of the induction step:

Question

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lublana lublana · Answer 1 · 2019-08-13T08:21:05+02:00

Answer with Step-by-step explanation:

We are given that

[tex]a_1=2[/tex]

[tex]a_n=a_{n-1}-3[/tex] for [tex] for [tex]n\geq2[/tex]

We have to prove that [tex]a_n=-3n+5[/tex] for all n equal and greater than 1

proof: Let P(n)=[tex]a_n=-3n+5[/tex] for all n equals to and greater than 1

We prove that for all n there exists Z,if n greater and equals to zero

Base case : The base case is n=1

Substitute n=1 then we get

P(1)=-3(1)+5=2

This is true for n=1

Induction step : suppose that P(n) is true for n=k where k greater than 1

Then P(k)=-3k+5

Now, we shall prove that for n=k+1, P(n) is true for n=k+1

P(k+1)=-3(k+1)+5

P(k+1)=-3k-3+5=-3k+5-3

[tex]P(k+1)=a_k-3[/tex]

[tex]a_{k+1}=a_k-3[/tex]

We are given that [tex]a_n=a_{n-1}-3[/tex] for n equal to and greater than 2

Therefore, P(n) is true for n=k+1

Hence, P(n)=-3n+5 for all n equal and greater than 1