Originally Posted by

**GrigOrig99** Can anyone help me confirm if I have solved this correctly?

Many thanks.

Q. (1 + ax)^{n} __>__ 1 + anx, for a > 0, x > 0 & n $\displaystyle \in \mathbb{N}$

**Attempt:** __Step 1:__ For n = 1...

(1 + ax)^{1} = 1 + ax & 1 + a(1)x = 1 + ax

Since 1 + ax __>__ 1 + ax, n is true.

__Step 2:__ Assume the statement is true for n = k, i.e. assume (1 + ax)^{k} __>__ 1 + akx.

We must now show that the statement is true for n = k + 1,

i.e. (1 + ax)^{k+1} __>__ 1 + a(k + 1)x = 1 + akx + ax

(1 + ax)^{k+1} = (1 + ax)(1 + ax)^{k} __>__ (1 + ax)(1 + akx)...((1 + ax)^{k} __>__ 1 + akx...assumed)

If (1 + ax)(1 + akx) __>__ 1 + akx + ax,