$2.$ Consider a proof of the following fact:

For all $,$ n $>=4,2^$n $>=$ n$^2$

$($provide brief explanation for your answer$)$

What should be proven in the base case?

a$.$ For n $=1,2^$n $>=$ n$^2$

b$.$ For n $=4,2^$n $>=$ n$^2$

c$.2^$k $>=$ k$^2$

d$.$ For every k $>4,$ if $2^$k $>=$ k$^2$ then $2^$k$+1>=($k$+1)^2$

answer: choice b $,$ since n$>=4,$ the smallest value for n$=4$

What should be proven in the inductive step?

a$.$ For n $=4,2^$n $>=$ n$^2$

b$.2^$k $>=$ k$^2$

c$.$ For every k $>4,$ if $2^$k $>=$ k$^2$ then $2^$k$+1>=($k$+1)^2$

Below is an argument for the inductive step. In which choice is the inductive

hypothesis used?

a$\mathrm{.}2^2$k$+1=2.2^$k

b$.2^2$k$+1>=2.$ k$^2$

c$.2^2$k$+1=$ k$^2+$k$^2$

d$.2^2$k$+1=$ k$^2+2$k $+2$k

answer is choice b

The assumption that $^2<=2^$ is true is the inductive hypothesis, and is used to

substitute $^2$ for $2^.$

$3.$ Prove that for any positive integer n$,6$ evenly divides $7^$n $1$ using induction.

Proof by induction on $.$