Question

State, by writing first four terms, the expansion of following, where \[\left| x \right| < 1\]
(i) \[{\left( {1 + x} \right)^{ - 4}}\]
(ii) \[{\left( {1 - x} \right)^{ - \dfrac{1}{3}}}\]
(iii) \[{\left( {1 - {x^2}} \right)^{ - 3}}\]
(iv) \[{\left( {1 + x} \right)^{ - \dfrac{1}{5}}}\]

Answer 1

Hint:Here in this question, we have to write the first four terms by expanding each binomial expression. Given a four binomial expression having finite power this each expression can be expand by using a binomial theorem i.e., \[{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}\,{a^{n - r}} \cdot {b^r}} \], where \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] to get the required solution.

Complete step by step answer:
Binomial Theorem is used to solve any binomial expressions in a simplest way. It gives an expression to calculate the expansion form of \[{\left( {a + b} \right)^n}\] for any positive integer n. The Binomial theorem is stated as: \[{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}\,{a^{n - r}} \cdot {b^r}} \] or
\[{\left( {a + b} \right)^n} = {\,^n}{C_0}\,{a^n} + {\,^n}{C_1}\,{a^{n - 1}}{b^1} + {\,^n}{C_2}\,{a^{n - 2}}{b^2} + ..... + {\,^n}{C_r}\,{a^{n - r}}{b^r} + ..... + {\,^n}{C_n}\,{b^n}\]-------(1)
If, \[\left| x \right| < 1\] then the binomial expansion is can be written as sum of an infinite series
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - n}} = \,1 + \dfrac{{ - n}}{1} \cdot {x^1} + \dfrac{{\left( { - n} \right)\left( { - n - 1} \right)}}{{1 \times 2}} \cdot {x^2} + \dfrac{{\left( { - n} \right)\left( { - n - 1} \right)\left( { - n - 2} \right)}}{{1 \times 2 \times 3}} \cdot {x^3} + ..... + \dfrac{{\left( { - n} \right)\left( { - n - 1} \right)\left( { - n - 2} \right)...\left( { - n - r + 1} \right)}}{{1 \times 2 \times ... \times r}} \cdot {x^r} + .....\infty \]
On simplification, we can written as
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - n}} = \,1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
                                                                                                                               -------(2)
The binomial expansion for \[{\left( {1 - x} \right)^{ - n}}\] is given by
\[ \Rightarrow \,\,{\left( {1 - x} \right)^{ - n}} = \,1 + nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} + \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
                                                                                                                               -------(3)
Consider the question: Given 4 binomial expression, we have to expand each expression up to first four terms:
(i) \[{\left( {1 + x} \right)^{ - 4}}\]
Now by using a expansion i.e., equation (2)
\[n = 4\]
On substituting, we have
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - 4}} = \,1 - 4x + \dfrac{{4\left( {4 + 1} \right)}}{{2!}} \cdot {x^2} - \dfrac{{4\left( {4 + 1} \right)\left( {4 + 2} \right)}}{{3!}} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - 4}} = \,1 - 4x + \dfrac{{4\left( 5 \right)}}{2} \cdot {x^2} - \dfrac{{4\left( 5 \right)\left( 6 \right)}}{6} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - 4}} = \,1 - 4x + \dfrac{{20}}{2} \cdot {x^2} - \dfrac{{120}}{6} \cdot {x^3} + .....\]
The first four terms of binomial \[{\left( {1 + x} \right)^{ - 4}}\] is
\[\therefore \,\,{\left( {1 + x} \right)^{ - 4}} = \,1 - 4x + 10{x^2} - 20{x^3} + .....\].

(ii) \[{\left( {1 - x} \right)^{ - \dfrac{1}{3}}}\]
Now by using a expansion i.e., equation (3)
\[n = \dfrac{1}{3}\]
On substituting, we have
\[ \Rightarrow \,\,{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} = \,1 + \dfrac{1}{3}x + \dfrac{{\dfrac{1}{3}\left( {\dfrac{1}{3} + 1} \right)}}{{2!}} \cdot {x^2} + \dfrac{{\dfrac{1}{3}\left( {\dfrac{1}{3} + 1} \right)\left( {\dfrac{1}{3} + 2} \right)}}{{3!}} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} = \,1 + \dfrac{x}{3} + \dfrac{{\dfrac{1}{3}\left( {\dfrac{4}{3}} \right)}}{2} \cdot {x^2} + \dfrac{{\dfrac{1}{3}\left( {\dfrac{4}{3}} \right)\left( {\dfrac{7}{3}} \right)}}{6} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} = \,1 + \dfrac{x}{3} + \dfrac{4}{{18}} \cdot {x^2} + \dfrac{{28}}{{162}} \cdot {x^3} + .....\]
The first four terms of binomial \[{\left( {1 - x} \right)^{ - \dfrac{1}{3}}}\] is
\[\therefore \,\,{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} = \,1 + \dfrac{x}{3} + \dfrac{{2{x^2}}}{9} + \dfrac{{14{x^3}}}{{81}} + .....\].

(iii) \[{\left( {1 - {x^2}} \right)^{ - 3}}\]
Now by using a expansion i.e., equation (3)
\[n = 3\]
On substituting, we have
\[ \Rightarrow \,\,{\left( {1 - {x^2}} \right)^{ - 3}} = \,1 + 3.{x^2} + \dfrac{{3\left( {3 + 1} \right)}}{{2!}} \cdot {x^4} + \dfrac{{3\left( {3 + 1} \right)\left( {3 + 2} \right)}}{{3!}} \cdot {x^6} + .....\]
\[ \Rightarrow \,\,{\left( {1 - {x^2}} \right)^{ - 3}} = \,1 + 3{x^2} + \dfrac{{3\left( 4 \right)}}{2} \cdot {x^4} + \dfrac{{3\left( 4 \right)\left( 7 \right)}}{6} \cdot {x^6} + .....\]
\[ \Rightarrow \,\,{\left( {1 - {x^2}} \right)^{ - 3}} = \,1 + 3{x^2} + \dfrac{{12}}{2} \cdot {x^4} + \dfrac{{84}}{6} \cdot {x^6} + .....\]
The first four terms of binomial \[{\left( {1 - {x^2}} \right)^{ - 3}}\] is
\[\therefore \,\,{\left( {1 - {x^2}} \right)^{ - 3}} = \,1 + 3{x^2} + 6{x^4} + 14{x^6} + .....\]

(iv) \[{\left( {1 + x} \right)^{ - \dfrac{1}{5}}}\]
Now by using a expansion i.e., equation (2)
\[n = - \dfrac{1}{5}\]
On substituting, we have
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - \dfrac{1}{5}}} = \,1 - \dfrac{1}{5}x + \dfrac{{\dfrac{1}{5}\left( {\dfrac{1}{5} + 1} \right)}}{{2!}} \cdot {x^2} - \dfrac{{\dfrac{1}{5}\left( {\dfrac{1}{5} + 1} \right)\left( {\dfrac{1}{5} + 2} \right)}}{{3!}} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - \dfrac{1}{5}}} = \,1 - \dfrac{x}{5} + \dfrac{{\dfrac{1}{5}\left( {\dfrac{6}{5}} \right)}}{2} \cdot {x^2} - \dfrac{{\dfrac{1}{5}\left( {\dfrac{6}{5}} \right)\left( {\dfrac{{11}}{5}} \right)}}{6} \cdot {x^3} + .....\]
\[ \Rightarrow \,\,{\left( {1 + x} \right)^{ - \dfrac{1}{5}}} = \,1 - \dfrac{x}{5} + \dfrac{6}{{50}} \cdot {x^2} - \dfrac{{66}}{{750}} \cdot {x^3} + .....\]
The first four terms of binomial \[{\left( {1 + x} \right)^{ - 4}}\] is
\[\therefore \,\,{\left( {1 + x} \right)^{ - \dfrac{1}{5}}} = \,1 - \dfrac{x}{5} + \dfrac{3}{{25}}{x^2} - \dfrac{6}{{125}}{x^3} + .....\].

Note: The four major binomial expansion is given as
\[{\left( {1 + x} \right)^n} = \,1 + nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} + \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
\[\Rightarrow {\left( {1 + x} \right)^{ - n}} = \,1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
 \[\Rightarrow {\left( {1 - x} \right)^n} = \,1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
\[\Rightarrow {\left( {1 - x} \right)^{ - n}} = \,1 + nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}} \cdot {x^2} + \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}} \cdot {x^3} + ..... + {\left( { - 1} \right)^r}\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)...\left( {n + r - 1} \right)}}{{r!}} \cdot {x^r} + .....\infty \]
Using these expansions, any kind of problem can be solved. The value of $x$ will depend on the question.