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

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Harish