Question

How do you use the binomial series to expand ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$?

Answer 1

The terms of the form ${{(a+b)}^{n}}$ are called binomial terms. To simplify these terms, we should know the binomial expansion. For the binomial terms of the form ${{\left( 1+x \right)}^{n}}$, where n is not a positive integer. These terms are expanded as,
$1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)\left( n-2 \right)}{3!}{{x}^{3}}+\dfrac{n(n-1)\left( n-2 \right)\left( n-3 \right)}{4!}{{x}^{4}}+......$. We will use this expansion formula to expand the given binomial term.

Complete step-by-step solution:
We are asked to expand the binomial term ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$. As the exponent is not an integer, this term is of the form ${{\left( 1+x \right)}^{n}}$, here we have $-x$ at the place of x and $n=\dfrac{1}{3}$.
We know that the expansion of the binomial term ${{\left( 1+x \right)}^{n}}$ is
$1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)\left( n-2 \right)}{3!}{{x}^{3}}+\dfrac{n(n-1)\left( n-2 \right)\left( n-3 \right)}{4!}{{x}^{4}}+......$
We can find the expansion of ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$ by replacing x by $-x$, and substituting $n=\dfrac{1}{3}$ in the above expansion formula, by doing this we get
$\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)}{2!}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)\left( \dfrac{1}{3}-2 \right)}{3!}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{1}{3}-1 \right)\left( \dfrac{1}{3}-2 \right)\left( \dfrac{1}{3}-3 \right)}{4!}{{\left( -x \right)}^{4}}+......$
Simplifying the numerators of the above expansion, we get
$\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2!}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{3!}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{4!}{{\left( -x \right)}^{4}}+......$
We know that the values of $1!,2!,3!,4!$ are 1, 2, 6, and 24 respectively. Substituting these values in the denominators of the above expression, we get
$\Rightarrow 1+\dfrac{1}{3}\left( -x \right)+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2}{{\left( -x \right)}^{2}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{6}{{\left( -x \right)}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{24}{{\left( -x \right)}^{4}}+......$
Simplifying the exponents, we get
$\Rightarrow 1-\dfrac{1}{3}x+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)}{2}{{x}^{2}}-\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)}{6}{{x}^{3}}+\dfrac{\dfrac{1}{3}\left( \dfrac{-2}{3} \right)\left( \dfrac{-5}{3} \right)\left( \dfrac{-8}{3} \right)}{24}{{x}^{4}}+......$
Finally, simplifying both numerators, and denominators of both of the above expression, we get
$\Rightarrow 1-\dfrac{1}{3}x-\dfrac{1}{9}{{x}^{2}}-\dfrac{5}{81}{{x}^{3}}-\dfrac{10}{243}{{x}^{4}}+......$
Thus, the binomial expansion of ${{\left( 1-x \right)}^{\dfrac{1}{3}}}$ is $1-\dfrac{1}{3}x-\dfrac{1}{9}{{x}^{2}}-\dfrac{5}{81}{{x}^{3}}-\dfrac{10}{243}{{x}^{4}}+......$.

Note: To solve the questions of binomial expansions, we should know the binomial expansions of different expressions. For a general binomial term of the form ${{(a+b)}^{n}}$, here n is a positive integer. The expansion formula is $\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}{{a}^{n-r}}{{b}^{r}}$. Here, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We can find the expansion of a binomial term with standard form using the summation form given above.