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

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: