Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$ (1+x)^n ≥ 1 + nx $$

So he checks for 1, and get:

$$ 1+x ≥ 1+x $$

Next for variable k:

$$ (1+x)^k ≥ 1 + kx $$

Then the book wanna prove:

$$ (1+x)^{k+1} ≥ 1 + (k + 1)x $$

And here is books proof:

$$ (1+x)^{k+1} = (1+x)^k (1+x) ≥ (1+kx)(1+x) $$ $$ = 1+(k+1)x + kx^2 ≥ 1 + (k + 1)x $$

Finished! Well... How did the book get this: $(1+kx)(1+x)$ in that last part? Sorry, I'm so confused. Sorry if this to easy to be here. Thanks for all help helping me understand it!

share|cite|improve this question

Your question is a great one, it is most welcome here! The answer is that, in a proof by induction, we first check the base case (here, it is $n=1$), and then, assuming the result is true for $n=k$, we prove that the result must also be true for $n=k+1$. In other words, we want to prove that $$\text{true for }n=k\implies\text{true for }n=k+1$$

Intuitively, this lets us say $$\begin{align} (\text{base case}) \qquad\qquad\qquad\qquad\qquad\qquad&\text{true for }n=1\qquad\checkmark\\ {\text{true for }n=1,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=2\qquad\checkmark\\ {\text{true for }n=2,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=3\qquad\checkmark\\ \vdots\end{align}$$

Thus, when we try to prove that the statement is true for $n=k+1$, i.e. $$(1+x)^{k+1} ≥ 1 + (k + 1)x,$$ we can use the assumption that the statement is true for $n=k$, i.e. $$(1+x)^k ≥ 1 + kx.$$ The reason why we have $$(1+x)^k (1+x) ≥ (1+kx)(1+x)$$ is that we are assuming $$(1+x)^k ≥ 1 + kx$$ is true, and then we multiply both sides by $(1+x)$.

share|cite|improve this answer
Actually, $n=0$ can be a base case as well :-) – Asaf Karagila Aug 21 '11 at 14:44

In your next to last line, you have $(1+x)^k,$ which you have assumed two lines above is greater than or equal to $1+kx$. So it made the substitution, using a $\ge$ sign. You might look at this answer, which has a detailed explanation of induction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.