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# verifying the inequality using induction

I have a question about verifying the inequality using induction.

2N + 1<or =  2^N,       N = 3, 4,....

So far, I have the basis
2(3) + 1 < or = 2^3
6 + 1 < or = 8
7 < or = 8

Trying to do the induction step and it is not making much sense. I am assuming that since I am using 3 in the basis, I am using n+3 in the inductive? The problem I think I am having the most trouble with is the power of.
This is what I have, although I am sure it is completely incorrect.

2(n+3) + 1 < or = 2^n+3
2n + 6 + 1 < or = 2^n+3
2n + 7 < or = 2^n+3

0
rcanter
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1 Solution

Commented:
Hi, you have proved it for N=3. Now, put N=N+1, and prove that

2(N+1) + 1 <=  2^(N+1)

Now, you have

2N + 1 <=  2^N

Multiply both sides by 2

4N + 2 <= 2^(N+1)

2(N+1) + 2N <= 2^(N+1)

Since N is positive, 2N is also positive. Also, N >= 3. Hence, 2N >= 6 and obviously 2N >= 1, so we can write,

2(N+1) + 1 <= 2(N+1) + 2N <= 2^(N+1)

Removing the intermediate term,

2(N+1) + 1 <= 2^(N+1)

Hence the proof

---
Harish
0

Author Commented:
U are awesome!!! Thanks so much!!!
One quick question, just so that I understand.  Even that I used S(3) for the basis, I am using S(n+1) for the inductive?
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Commented:
rcanter, yes. You used S(3) because it is given in the problem itself that N = 3,4...

As I told in the previous question, you need to prove it that the (in)equality holds for atleast one value, which in case if it is not given, will be taken generally as 1
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Author Commented:
I am not understanding how this could be proven.
2N + 1 <=  2^N

Multiply both sides by 2     ***why am i multiplying by 2?***

4N + 2 <= 2^(N+1)

2(N+1) + 2N <= 2^(N+1)     **why wouldnt this be 2(n+1) + 1? how did 2n figure in?**

Since N is positive, 2N is also positive. Also, N >= 3. Hence, 2N >= 6 and obviously 2N >= 1, so we can write,

2(N+1) + 1 <= 2(N+1) + 2N <= 2^(N+1)

Removing the intermediate term,

2(N+1) + 1 <= 2^(N+1)

Hence the proof
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Commented:
you need to get N+1 in either side and then prove the other side
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