Please help me solve this question:
Prove that 2^n > 3n for all positive integers n `>=` 4
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This can be solved by mathematical induction.
It has given that n>=4.
So the lowest value of n=4
`2^4 = 16`
`3xx4 = 12`
So when n = 4 the result is true.
Let us assume when n = p where p>4 the result is true.
Now we have to prove that when `n = p+1` the result is true.
What we have to show that is `2^(p+1)>3(p+1)` .
But what we have obtained is `2^(p+1)>6p` .
So if we can prove `6p>3(p+1)` then we have say;
So ultimately we will endup with `2^(p+1)>3(p+1)`
Let us assume `6p>3(p+1)`
`6p-3(p+1) > 0`
`3(p-1) > 0`
We know that `p>4` . Then `(p-1)>0` and `3(p-1) > 0` .
This means our assumption is correct.
Now we can say `2^(p+1)>3(p+1)`
So for n = p+1 the result is correct.
Therefore from mathematical induction for all `n>=4` ;
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