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

Many thanks.

Q. (1 + 2x)

^{n} __>__ 1 + 2nx, for x > 0 & n

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

(1 + 2x)

^{1} = 1 + 2x '&' 1 + 2(1)x = 1 + 2x

Since 1 + 2x

__>__ 1 + 2x, n is true.

Step 2: Assume the statement is true for n = k, i.e. assume (1 + 2x)

^{k} __>__ 1 + 2kx,

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

i.e. (1 + 2x)

^{k+1} __>__ 1 + 2(k + 1)x = 1 + 2kx + 2x

(1 + 2x)

^{k+1} = (1 + 2x)

^{1}(1 + 2x)

^{k} __>__ (1 + 2x)(1 + 2kx)...((1 + 2x)

^{k} __>__ 1 + 2kx...assumed)

If (1 + 2x)(1 + 2kx)

__>__ 1 + 2kx + 2x then (1 + 2x)

^{k+1} __>__ 1 + 2kx + 2x

if (1 + 2x)(1 + 2kx)

__>__ 1 + 2kx + 2x

if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x)

__>__ 0

if 2kx

^{2} + 2kx + 2x + 1 - 1 - 2kx - 2x

__>__ 0

if 2kx

^{2} __>__ 0

if x

^{2}[2k]

__>__ 0...true when x > 0 & n

__>__ 1

(1 + 2x)

^{k+1} __>__ 1 + 2kx + 2x

Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n

__>__ 1, n