demonstration by induction: $(1+a)^n ≥1+an$

Demonstrate by induction: $(1+a)^n ≥1+an$ is true, given a real number $a$, for any $n ∈ \mathbb N$. With $a > 0$

I need to demostre this using the induction principle. My doubt is in the second part of the demonstration.

This is what I have:

In order to demonstrate the predicate these two points must be true:

$P(1)$ must be True.
$∀ n ∈ \mathbb N , P(n) => P(n+1)$ is true.

We demonstrate the first point:

$(1+a)^1 ≥ 1+a*1$, $1+a ≥ 1+a$ So it is true

Now, the second part is where I have the problem. I do not know what to do. I understand the theory but I don't know how to apply it.

I tried this:

$(1+a)^{n+1} ≥ 1+a(n+1)$

But I don't see that as useful.

Any tips?

Solution 1:

Ross has pretty much given you all the hints necessary to answer this particular question, so let me instead give you some general advice on proofs by induction.

In doing the induction, especially at first and until you get very comfortable with induction, be very explicit. Label the base Base. Label the inductive step as "Inductive step." State explicitly what the Induction Hypothesis is, state explicitly what you want to prove. Use this as a way to organize your thoughts and have the information at the ready.

Second: The key to most proofs by induction is to take the case you need to prove, and to somehow reduce it to "the previous case plus something extra"; then one applies the inductive hypothesis to simplify/answer the part that is the previous case, and then deal with the "something extra."

Here's an example different from the one at hand, so you can see what I mean.

Consider the following:

Prove that for all natural numbers $n$, $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}.$$

Proof. We proceed by indution on $n$.

Base. We prove the statement for $n=1$: indeed, $1\cdot 2 = \frac{1(2)(3)}{3}$.

Inductive step.

Induction Hypothesis. We assume the result holds for $k$. That is, we assume that $$ 1\cdot 2 + 2\cdot 3 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}$$ is true.

To prove: We need to show that the result holds for $k+1$, that is, that $$ 1 \cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}.$$

(Now comes the point where we take $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)$ and try to think of it as "the $k$-case plus something extra").

\begin{align*} &{1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)}\\ \quad &= \Bigl(1\cdot 2+ 2\cdot 3+\cdots + k(k+1)\Bigr) + (k+1)(k+2) &&\mbox{(associativity of $+$)}\\ \quad&= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) &&\mbox{(This is the induction hypothesis!)}\\ &= \frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3} &&\mbox{(Just algebra)}\\ &= \frac{(k+1)(k+2)\bigl(k+3\bigr)}{3} &&\mbox{(factor out $(k+1)(k+2)$)}\\ &= \frac{(k+1)\bigl( (k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}. \end{align*} Thus, $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}$. This proves the inductive step.

Since the statement holds for $n=1$, and if it holds for $k$ then it holds for $k+1$, then by mathematical induction we conclude that $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}$$ for all natural numbers $n$. QED

So: be very explicit and clear, to help organize your thoughts. As you get more comfortable with induction, you can leave off makign explicit statement of what you need to prove, etc., but until you are very comfortable, better to be explicit than to be confused.

Solution 2:

Hint: Now you assume that $(1+a)^n \ge 1 + an$ and need to prove that $(1+a)^{(n+1)} \ge 1 + a(n+1)$. So $(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge $ what?

Added:$(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge (1+an)(1+a)$ by the inductive hypothesis $(1+an)(1+a)=1+a(n+1)+a^2n \ge 1+a(n+1)$ So, as Arturo showed in his example, we have proven that if the relation is true for $n$, it is also true for $n+1$. Given that you verified the base case of $n=1$, it is true for all $n$