For `a gt 1`, prove that `(1+a)^(n) ge (1+an)` for all natural numbers `n`.

Explanation

When n=1 LHS=(1+a) and RHS =(1+a)
Hence `(1+a)^(n) ge (1+na)` is true, when n=1
`therefore P(1)` is true. (a)
Let P(k) be true `Rightarrow (1+a)^(k) ge (1+ka)`…..(i)
To prove P(k+1) is true, i.e. `(1+a)^(k+1) ge 1+(k+1)a=x("say")` ......(ii) `"from "(i), (1+a)^(k+1) ge (1+ka)(1+a)y(say)`
Now, (y-x)=`(1+ka) +a+ka^(2)-1-ka-a)`
`=1+ka+a+ka^(2)-1-1ka-a`
`=ka^(2) gt 0 therefore y gt x`
From (iii) and (iv), we get
`(1+a)^(k+1) ge 1,+(k+1)a`
Hence, P(k+1) is true whenever P(k) is true......(b)
From (a) and (b) by the principle of mathematical induction, it follows that P(n) is true for all natural number n.