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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)

Fill in the following blanks to prove that n 21 n lt 2n n1 lt 2n1 is Box 3 Options True False Next assume that Box 4 Options 1 lt 21 k 1 lt 2k1 k lt 2k as we at class=

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wegnerkolmp2741o

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)

Alfpfeu

Answer:

Step-by-step explanation:

Hello, please consider the following.

First, assume that n equals [tex]\boxed{1}[/tex]. Therefore, [tex]\boxed{1<2^1}[/tex] is [tex]\boxed{\text{True}}[/tex]

Next, assume that [tex]\boxed{k<2^k}[/tex], as we attempt to prove [tex]\boxed{k+1<2^{k+1}}[/tex]

Since .... Therefore, we can conclude that [tex]\boxed{k+1<2^{k+1}}[/tex]

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 [tex]n\geq 1[/tex], we have [tex]n<2^n[/tex].

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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