Hi,

I had this induction problem which i couldn't figure out how to solve.

for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

The solution looked like this:
$\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\ \geq ~(1+x){\color{red}(1+kx)} \\ =~1+kx+x+kx^2\\ \geq~1+kx+x\\ =~1+(k+1)x \end{array}$

$\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
How can the stuff marked in red be valid?

2. That's precisely the induction step -- you assumed that $\displaystyle (1+x)^n \geq 1 + nx$, so you get $\displaystyle (1+x)(1+x)^k \geq (1+x)(1 + kx)$

And the rest follows easily..

3. Originally Posted by Jones
Hi,

I had this induction problem which i couldn't figure out how to solve.

for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

The solution looked like this:
$\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\ \geq ~(1+x){\color{red}(1+kx)} \\ =~1+kx+x+kx^2\\ \geq~1+kx+x\\ =~1+(k+1)x \end{array}$

$\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
How can the stuff marked in red be valid?
This is a rather well known thing called Bernoulli's Inequality, and it MUST be that $\displaystyle x>-1$

Tonio

4. Originally Posted by Jones
Hi,

I had this induction problem which i couldn't figure out how to solve.

for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

The solution looked like this:
$\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\ \geq ~(1+x){\color{red}(1+kx)} \\ =~1+kx+x+kx^2\\ \geq~1+kx+x\\ =~1+(k+1)x \end{array}$

$\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
How can the stuff marked in red be valid?

at that part you bolded they are replacing the (1+x)^k with what we know.

we know that $\displaystyle (1+x)^n \geq 1+nx$.

at the part you bolded, $\displaystyle (1+x)(1+x)^k$ at this part we replace$\displaystyle (1+x)^k$ with $\displaystyle (1+kx)$ .

leaving us with $\displaystyle (1+x)(1+kx)$.