Question

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

Answer 1

Hint: The given question requires us to find the binomial expansion of the given binomial expression \[{\left( {1 + x} \right)^{ - 3}}\]. We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete answer:
We have to find the binomial expansion of \[{\left( {1 + 4x} \right)^{ - 3}}\] . So, using the binomial theorem, the binomial expansion of \[{\left( {1 + x} \right)^n}\] is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + ....$
So, the binomial expansion of \[{\left( {1 + x} \right)^{ - 3}}\] can be calculated by the above expression by substituting in the correct value of n and x.
Now, we compare the expression \[{\left( {1 + x} \right)^{ - 3}}\] with \[{\left( {1 + x} \right)^n}\]. So, we get the value of n as $ - 3$.
So, we have, \[ = 1 + \left( { - 3} \right)\left( x \right) + \dfrac{{\left( { - 3} \right)\left( {\left( { - 3} \right) - 1} \right)}}{{2!}}{\left( x \right)^2} + \dfrac{{\left( { - 3} \right)\left( {\left( { - 3} \right) - 1} \right)\left( {\left( { - 3} \right) - 2} \right)}}{{3!}}{\left( x \right)^3} + \dfrac{{\left( { - 3} \right)\left( {\left( { - 3} \right) - 1} \right)\left( {\left( { - 3} \right) - 2} \right)\left( {\left( { - 3} \right) - 3} \right)}}{{4!}}{\left( x \right)^4} + ....\] So, simplifying the expression further, we get,
\[ = 1 + \left( { - 3} \right)\left( x \right) + \dfrac{{\left( { - 3} \right)\left( { - 4} \right)}}{{2!}}{\left( x \right)^2} + \dfrac{{\left( { - 3} \right)\left( { - 4} \right)\left( { - 5} \right)}}{{3!}}{\left( x \right)^3} + \dfrac{{\left( { - 3} \right)\left( { - 4} \right)\left( { - 5} \right)\left( { - 6} \right)}}{{4!}}{\left( x \right)^4} + ....\]
So, we can evaluate first few terms of the binomial expansion,
\[ = 1 + \left( { - 3} \right)\left( x \right) + \dfrac{{12}}{2}{\left( x \right)^2} + \dfrac{{ - 60}}{6}{\left( x \right)^3} + \dfrac{{360}}{{24}}{\left( x \right)^4} + ....\]
Simplifying the expression and opening the brackets, we get,
\[ = 1 + \left( { - 3x} \right) + 6{\left( x \right)^2} - 10{\left( x \right)^3} + 15{\left( x \right)^4} + ....\]
\[ = 1 - 3x + 6{x^2} - 10{x^3} + 15{x^4} + ....\]
Hence, the expression \[{\left( {1 + x} \right)^{ - 3}}\] can be simplified as \[\left( {1 - 3x + 6{x^2} - 10{x^3} + 15{x^4} + ....} \right)\] using binomial theorem.

Note:
The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. Binomial series with negative indices should be solved using the formula $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + ....$. BODMAS rule should be followed while simplifying the expression and finding the final answer.