Problem source

Use the first four terms of the binomial expansion of  $(1-1/50)^{1/2}$, writing $1/50 = 2/100$ to simplify the calculation,  to derive the approximation
$\sqrt 2 \approx 1.414214$.
Calculate similarly an approximation to the cube root of 2 to six decimal places by considering $(1+N/125)^a$, where $a$ and $N$ are suitable numbers.
[You need not justify the accuracy of your approximations.]

Solution source

\begin{align*}
&& (1-1/50)^{1/2} &= 1 + \frac12 \cdot \left ( -\frac1{50} \right) + \frac1{2!} \frac12 \cdot \left ( -\frac12 \right)\cdot \left ( -\frac1{50} \right)^2 + \frac1{3!} \frac12 \cdot \left ( -\frac12 \right) \cdot \left ( -\frac32 \right)\cdot \left ( -\frac1{50} \right)^3 + \cdots \\
&&&=1-\frac{1}{100} - \frac12 \frac1{10000}  -\frac12 \frac1{1000000} +\cdots \\
&&&= 0.9899495 + \cdots \\
\Rightarrow && \frac{7\sqrt{2}}{10} &\approx 0.9899495 \\
\Rightarrow && \sqrt{2} &\approx \frac{9.899495}{7} \\
&&&\approx 1.414214
\end{align*}

\begin{align*}
&& (1 + 3/125)^{1/3} &= \frac{\sqrt[3]{125+3}}{5} \\
&&& = \frac{8\sqrt[3]{2}}{10} \\
&& (1 + 3/125)^{1/3} &= 1 + \frac13 \left ( \frac{3}{125} \right) + \frac1{2!} \cdot \frac{1}{3} \cdot \left ( -\frac23\right) \left ( \frac{3}{125}\right)^2 +\cdots \\
&&&= 1+ \frac{8}{1000} - \frac{64}{1000000} \\
&&&= 1.007936 \\
\Rightarrow && \sqrt[3]{2} &= \frac{10.07936}{8} \\
&&&= 1.259920
\end{align*}