Show by using the principle of mathematical induction that for all natural number `n gt 2, 2^(n) gt 2n+1`

Explanation

Let `P(n): 2^(n) gt 2n+1`, where `n gt 2` .
When n=3, LHS=`2^(3)=8 and RHS=2xx3+1=7`

clearly `8 gt 7` therefore `P(3)` is true. (a)
Let P(k) be true
In (i), multiplying both sides by 2, we get
`2^(k+1) gt 4k+2…… (iii)`
Now, (4k+2)-(2k+3)=`2k-1 gt 0`
`Rightarrow 4k+2 gt 2k+3 ....(iv)`
From (iii) and (iv), we get
`2^(k+1) gt 2k+3`
Hence P(k+1) is true whenever P(k) is true,
From (a) and (b), it follows that P(n) is true for all natual number n.