Question

How do you use the binomial series to expand $\dfrac{1}{{{{\left( {2 + x} \right)}^5}}}$?

Answer 1

In this type of problem, first recast the expression that we want to expand binomially in the form ${\left( {1 + x} \right)^n}$ and then use the binomial expansion for ${\left( {1 + x} \right)^n}$ to expand it by putting $n = - 5$ and $x = \dfrac{x}{2}$. Then, we will get the binomial expansion of $\dfrac{1}{{{{\left( {2 + x} \right)}^5}}}$ for $\left| x \right| < 2$.
For $\left| x \right| > 2$, substitute $n = - 5$ and $x = \dfrac{2}{x}$ in the binomial expansion for ${\left( {1 + x} \right)^n}$ .
Formula used:
For an arbitrary $n$, the binomial expansion for ${\left( {1 + x} \right)^n}$ is given by
$1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ...$

Complete step-by-step solution:
Since, we know that for an arbitrary $n$, the binomial expansion for ${\left( {1 + x} \right)^n}$ is given by
$1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ...$
So, we first recast the expression that we want to expand binomially in the form ${\left( {1 + x} \right)^n}$.
Here we first use
$\dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{{2^5}}}{\left( {1 + \dfrac{x}{2}} \right)^{ - 5}}$
and then use the binomial expansion for $n = - 5$ to get
$\dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{{2^5}}}\left( {1 + \left( { - 5} \right)\dfrac{x}{2} + \dfrac{{\left( { - 5} \right)\left( { - 6} \right)}}{{2!}}{{\left( {\dfrac{x}{2}} \right)}^2} + \dfrac{{\left( { - 5} \right)\left( { - 6} \right)\left( { - 7} \right)}}{{3!}}{{\left( {\dfrac{x}{2}} \right)}^3} + ...} \right)$
Now, simplify the terms on the right side of the equation by multiplication.
$\Rightarrow \dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{32}}\left( {1 - \dfrac{5}{2}x + \dfrac{{15}}{4}{x^2} - \dfrac{{35}}{8}{x^3} + ...} \right)$
Of course, this infinite series only converges for $\left| {\dfrac{x}{2}} \right| < 1$, i.e., $- 2 < x < 2$.
If $\left| {\dfrac{x}{2}} \right| > 1$, this series will diverge - rendering it useless! What saves the day is that in this case, $\left| {\dfrac{2}{x}} \right| < 1$ and so, we can still get a binomial expansion, but this time we must start from
$\dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{{2^5}}}{\left( {1 + \dfrac{2}{x}} \right)^{ - 5}}$
and continue with the binomial expansion for $n = - 5$ to get
$\dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{{2^5}}}\left( {1 + \left( { - 5} \right)\dfrac{2}{x} + \dfrac{{\left( { - 5} \right)\left( { - 6} \right)}}{{2!}}{{\left( {\dfrac{2}{x}} \right)}^2} + \dfrac{{\left( { - 5} \right)\left( { - 6} \right)\left( { - 7} \right)}}{{3!}}{{\left( {\dfrac{2}{x}} \right)}^3} + ...} \right)$
Now, simplify the terms on the right side of the equation by multiplication.
$\Rightarrow \dfrac{1}{{{{\left( {2 + x} \right)}^5}}} = \dfrac{1}{{32}}\left( {1 - \dfrac{{10}}{x} + \dfrac{{60}}{{{x^2}}} - \dfrac{{280}}{{{x^3}}} + ...} \right)$
Final solution: Hence, the binomial expansion $\dfrac{1}{{{{\left( {2 + x} \right)}^5}}}$ is $\dfrac{1}{{32}}\left( {1 - \dfrac{5}{2}x + \dfrac{{15}}{4}{x^2} - \dfrac{{35}}{8}{x^3} + ...} \right)$ for $\left| x \right| < 2$ and $\dfrac{1}{{32}}\left( {1 - \dfrac{{10}}{x} + \dfrac{{60}}{{{x^2}}} - \dfrac{{280}}{{{x^3}}} + ...} \right)$ for $\left| x \right| > 2$.

Note: One important point to remember is that if the index $n$ is not a positive integer, the binomial expansion leads to an infinite series. In such a situation, it is important to keep in mind that the infinite binomial series for ${\left( {1 + x} \right)^n}$ converges only if $\left| x \right| < 1$.