Question

How do you use the binomial series to expand ${{\left( 1+x \right)}^{\dfrac{1}{2}}}$?

Answer 1

Binomial series is {{(x+y)}^{n}}=\underset{k=0}{\overset{n}{\mathop \sum }}\,(\begin{matrix} n \\\ k \\\ \end{matrix}){{x}^{n-k}}{{y}^{k}}=\underset{k=0}{\overset{n}{\mathop \sum }}\,(\begin{matrix} n \\\ k \\\ \end{matrix}){{x}^{k}}{{y}^{n-k}}
Where, $(\begin{matrix} n \\\ k \\\ \end{matrix})$ is the binomial coefficient or in general language we say it as $^{n}{{C}_{r}}$ where $C$ is “combination”. But to use this $n$ and $k$ must be positive if not then we use the derived formula for fractional or negative values of $n$. Now substitute the value of $n$ in the formula and pen down others as it as and this series continues to infinity but as soon as the number of terms increases and due to the fractional value of $n$ the term value decreases and will be negligible so in this we have considered first two terms.
However, with a fractional exponent, the formula is:
${{(1+x)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+...$ which continues to infinity.

Complete step by step solution:
As we know that the binomial series for the fractional component is ${{(1+x)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+...$ which continues to infinity
On Comparing it with the given problem-
$n=\dfrac{1}{2}$
Now substitute this value in above equation
$\Rightarrow {{(1+x)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}x+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)}{2!}{{x}^{2}}+\dfrac{\dfrac{1}{2}\left( \dfrac{1}{2}-1 \right)\left( \dfrac{1}{2}-2 \right)}{3!}{{x}^{3}}...$
Now simplifying
$\Rightarrow {{(1+x)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}x+\dfrac{-1}{8}{{x}^{2}}+\dfrac{3{{x}^{3}}}{48}...$
On simplifying the above expression
We get,
${{(1+x)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}x+\dfrac{-1}{8}{{x}^{2}}+\dfrac{3{{x}^{3}}}{48}...$

Hence the binomial expansion is ${{(1+x)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}x+\dfrac{-1}{8}{{x}^{2}}+\dfrac{3{{x}^{3}}}{48}...$

Note:
During solving this by binomial series note that the exponent power of the whole bracket is neither fractional nor negative otherwise we cannot use $^{n}{{C}_{r}}$ but we can use the above formula which we have used in this question which is only the derived formula from $^{n}{{C}_{r}}$ means we just have to put the values of $n$ and $r$ which can be put in derived formula as these can be fractional but also negative too. Those are the first four terms of the infinite series, and they’re valid if:
[\dfrac{-1}{2}