Using $(1+x)^k \approx 1+kx$ to approximate? 
Use the approximation $(1+x)^k \approx 1+kx$ to estimate $(1.0003)^{26}$ and $\sqrt[4]{1.006}$. 

I know how to solve this step-by-step, but I don't understand what I'm doing exactly: why does $(1+x)^k \approx 1+kx$?  
$x^k$ is $x \times x \times x \times \dots x\;\;$ a certain $k$ number of times. Where do they get $kx$ from? 
 A: If you expand $(1+x)^k$ you get $1^k+k1^{k-1}x+{k \choose 2}1^{k-2}x^2+\dots =1+kx +\frac 12k(k-1)x^2+\dots $  If $kx \ll1$ the following terms are small.
A: One can expand the function $(1 + x)^{\alpha}$ in a Taylor series called the binomial series. It takes on the form
$$(1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!} x^2 + \frac{\alpha (\alpha - 1)(\alpha - 2)}{3!} x^3 + \dots$$
Now if $x$ is small and $\alpha$ isn't too large relative to $x$, all the high-order terms can be neglected to give the first order approximation you're asking about.
A: First define the function $f(x)=(1+x)^n$ then using the Maclaurin series expansion defined as
\begin{equation}
f(x)=P_n(x)=\sum^n_{k=0}\frac{f^{(k)}(0)}{k!}x^k
\end{equation}
where $P_n(x)$ is a polynomial of degree $n$. Now we can calculate the first two derivatives of $f(x)$ 
\begin{align}
\left.\frac{df}{dx}\right|_{x=0}&=n(1+x)^{n-1}=n \\
\left.\frac{d^2f}{dx^2}\right|_{x=0}&=n(n-1)(1+x)^{n-2}=n(n-1)
\end{align}
Using the definition of the Maclaurin series the function can be written as
$$
f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{f^{(3)}(\xi)}{3!}x^3
$$
Where $\xi$ is the error term in the series expansion. Now when $|x|\ll1,\ $ $x^2$  and $x^3$ will be negligible and $f(x)$ will be very well approximated by the first two terms of the Maclaurin series giving:
$$
f(x)\approx1+nx
$$
A: Check out Bernoulli's Inequality (see http://en.wikipedia.org/wiki/Bernoulli%27s_inequality) which uses the binomial theorem.
