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:

Answer with Step-by-step explanation:

We are given that

for

We have to prove that for all n equal and greater than 1

proof: Let P(n)= 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

We are given that 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

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9