Question

How do you use the Binomial Theorem to expand ${\left( {1 + x} \right)^{ - 1}}$ ?

Answer 1

Here they have asked to expand the given function using binomial theorem. So we need to make use of the binomial theorem series expansion given by: ${\left( {1 + y} \right)^n} = \sum\limits_{k = 0}^\infty {\left( {_k^n} \right){y^k} = } 1 + ny + \dfrac{{n(n - 1)}}{{2!}}{y^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{y^3} + .......$. By using this series we can expand the given function.

Complete step by step answer:
In this question they have asked to expand the given function which is ${\left( {1 + x} \right)^{ - 1}}$ by using the binomial theorem. So first we need to know the binomial theorem series, which is given by: ${\left( {1 + y} \right)^n} = \sum\limits_{k = 0}^\infty {\left( {_k^n} \right){y^k} = } 1 + ny + \dfrac{{n(n - 1)}}{{2!}}{y^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{y^3} + .......$.
Therefore, by comparing the given function ${\left( {1 + x} \right)^{ - 1}}$ with left hand side of the binomial series ${\left( {1 + y} \right)^n}$ we can say that $y = x$ and $n = - 1$.
Now substitute $y = x$ and $n = - 1$ in the above binomial theorem formula or the series, to get the expanded form of the given function ${\left( {1 + x} \right)^{ - 1}}$.
On substitution, we get
${\left( {1 + x} \right)^{ - 1}} = 1 + ( - 1)(x) + \dfrac{{( - 1)( - 1 - 1)}}{{2!}}{x^2} + \dfrac{{( - 1)( - 1 - 1)( - 1 - 2)}}{{3!}}{x^3} + .......$
Now, simplify the above expression in order to get the simplified version of it.
${\left( {1 + x} \right)^{ - 1}} = 1 + - x + \dfrac{2}{{2!}}{x^2} + \dfrac{{( - 1)( - 2)( - 3)}}{{3!}}{x^3} + .......$
Factorial of a number is nothing but the product of all the numbers which is less than and equal to the given number. Hence the factorial is of the numbers in the above expression is written as below,
${\left( {1 + x} \right)^{ - 1}} = 1 + - x + \dfrac{2}{{2 \times 1}}{x^2} + \dfrac{{ - 6}}{{3 \times 2 \times 1}}{x^3} + .......$
${\left( {1 + x} \right)^{ - 1}} = 1 + - x + \dfrac{2}{2}{x^2} + \dfrac{{ - 6}}{6}{x^3} + .......$
Now we have common terms at the right hand side both in numerator and denominator which get cancelled for the simplification purpose.
Therefore, we get
${\left( {1 + x} \right)^{ - 1}} = 1 + - x + {x^2} - {x^3} + .......$

Hence the expanded version of ${\left( {1 + x} \right)^{ - 1}}$ by using binomial series is ${\left( {1 + x} \right)^{ - 1}} = 1 + - x + {x^2} - {x^3} + .......$.

Note:
Whenever we have to extend any of the functions, we need to know the related formulas or the expression ten only we can get the correct answer. Otherwise, we cannot get the correct answer. So we need to remember the binomial series to expand the given function.