Question

Fill in the following blanks to prove that n 2^1 n < 2^n n+1 < 2^(n+1) is Box 3 Options: True | False Next, assume that Box 4 Options: 1 < 2^1 k + 1 < 2^(k+1) k < 2^k as we attempt to prove Box 5 Options: k < 2^k k + 1 < 2^(k+1) 2 < 2^1 Therefore, we can conclude that Box 6 Options: k < 2^k k + 1 < 2^(k+1) 2^1 < 2^k k + 2 < 2^(k+2)

Answer 1

Answer:

Step-by-step explanation:

Hello, please consider the following.

First, assume that n equals $\boxed{1}$. Therefore, $\boxed{1$ is $\boxed{\text{True}}$

Next, assume that $\boxed{k$, as we attempt to prove $\boxed{k+1$

Since .... Therefore, we can conclude that $\boxed{k+1$

The choice for the last box is confusing. Based on your feedback, we can assume that we are still in the step 2 though.

And the last step which is not included in your question is the conclusion where we can say that we prove that for any integer $n\geq 1$, we have $n$.

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer 2

Answer:

see below

Step-by-step explanation:

n < 2^n

First let n=1

1 < 2^1

1 <2  This is true

Next, assume that

(k) < 2^(k)

as we attempt to prove that

(k+1) < 2^(k+1)

.

.

.

Therefore we can conclude that

k+1 < 2^(k+1)