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Serga [27]
2 years ago
9

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 Bo

x 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)

Mathematics
2 answers:
larisa [96]2 years ago
5 0

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

miskamm [114]2 years ago
4 0

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)

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