Solveeit Logo
Login

Question

Question: State, by writing first four terms, the expansion of following, where \[\left| x \right| < 1\] (i)...

State, by writing first four terms, the expansion of following, where ∣x∣<1\left| x \right| < 1
(i) (1+x)−4{\left( {1 + x} \right)^{ - 4}}
(ii) (1−x)−13{\left( {1 - x} \right)^{ - \dfrac{1}{3}}}
(iii) (1−x2)−3{\left( {1 - {x^2}} \right)^{ - 3}}
(iv) (1+x)−15{\left( {1 + x} \right)^{ - \dfrac{1}{5}}}

Explanation

Solution

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., (a+b)n=∑r=0nnCr an−r⋅br{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}\,{a^{n - r}} \cdot {b^r}} , where nCr=n!(n−r)!r!^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 (a+b)n{\left( {a + b} \right)^n} for any positive integer n. The Binomial theorem is stated as: (a+b)n=∑r=0nnCr an−r⋅br{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}\,{a^{n - r}} \cdot {b^r}} or
(a+b)n= nC0 an+ nC1 an−1b1+ nC2 an−2b2+.....+ nCr an−rbr+.....+ nCn bn{\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, ∣x∣<1\left| x \right| < 1 then the binomial expansion is can be written as sum of an infinite series
⇒  (1+x)−n= 1+−n1⋅x1+(−n)(−n−1)1×2⋅x2+(−n)(−n−1)(−n−2)1×2×3⋅x3+.....+(−n)(−n−1)(−n−2)...(−n−r+1)1×2×...×r⋅xr+.....∞\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
⇒  (1+x)−n= 1−nx+n(n+1)2!⋅x2−n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞\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 (1−x)−n{\left( {1 - x} \right)^{ - n}} is given by
⇒  (1−x)−n= 1+nx+n(n+1)2!⋅x2+n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞\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) (1+x)−4{\left( {1 + x} \right)^{ - 4}}
Now by using a expansion i.e., equation (2)
n=4n = 4
On substituting, we have
⇒  (1+x)−4= 1−4x+4(4+1)2!⋅x2−4(4+1)(4+2)3!⋅x3+.....\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} + .....
⇒  (1+x)−4= 1−4x+4(5)2⋅x2−4(5)(6)6⋅x3+.....\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} + .....
⇒  (1+x)−4= 1−4x+202⋅x2−1206⋅x3+.....\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 (1+x)−4{\left( {1 + x} \right)^{ - 4}} is
∴  (1+x)−4= 1−4x+10x2−20x3+.....\therefore \,\,{\left( {1 + x} \right)^{ - 4}} = \,1 - 4x + 10{x^2} - 20{x^3} + ......

(ii) (1−x)−13{\left( {1 - x} \right)^{ - \dfrac{1}{3}}}
Now by using a expansion i.e., equation (3)
n=13n = \dfrac{1}{3}
On substituting, we have
⇒  (1−x)−13= 1+13x+13(13+1)2!⋅x2+13(13+1)(13+2)3!⋅x3+.....\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} + .....
⇒  (1−x)−13= 1+x3+13(43)2⋅x2+13(43)(73)6⋅x3+.....\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} + .....
⇒  (1−x)−13= 1+x3+418⋅x2+28162⋅x3+.....\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 (1−x)−13{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} is
∴  (1−x)−13= 1+x3+2x29+14x381+.....\therefore \,\,{\left( {1 - x} \right)^{ - \dfrac{1}{3}}} = \,1 + \dfrac{x}{3} + \dfrac{{2{x^2}}}{9} + \dfrac{{14{x^3}}}{{81}} + ......

(iii) (1−x2)−3{\left( {1 - {x^2}} \right)^{ - 3}}
Now by using a expansion i.e., equation (3)
n=3n = 3
On substituting, we have
⇒  (1−x2)−3= 1+3.x2+3(3+1)2!⋅x4+3(3+1)(3+2)3!⋅x6+.....\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} + .....
⇒  (1−x2)−3= 1+3x2+3(4)2⋅x4+3(4)(7)6⋅x6+.....\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} + .....
⇒  (1−x2)−3= 1+3x2+122⋅x4+846⋅x6+.....\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 (1−x2)−3{\left( {1 - {x^2}} \right)^{ - 3}} is
∴  (1−x2)−3= 1+3x2+6x4+14x6+.....\therefore \,\,{\left( {1 - {x^2}} \right)^{ - 3}} = \,1 + 3{x^2} + 6{x^4} + 14{x^6} + .....

(iv) (1+x)−15{\left( {1 + x} \right)^{ - \dfrac{1}{5}}}
Now by using a expansion i.e., equation (2)
n=−15n = - \dfrac{1}{5}
On substituting, we have
⇒  (1+x)−15= 1−15x+15(15+1)2!⋅x2−15(15+1)(15+2)3!⋅x3+.....\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} + .....
⇒  (1+x)−15= 1−x5+15(65)2⋅x2−15(65)(115)6⋅x3+.....\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} + .....
⇒  (1+x)−15= 1−x5+650⋅x2−66750⋅x3+.....\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 (1+x)−4{\left( {1 + x} \right)^{ - 4}} is
∴  (1+x)−15= 1−x5+325x2−6125x3+.....\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
(1+x)n= 1+nx+n(n+1)2!⋅x2+n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞{\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
⇒(1+x)−n= 1−nx+n(n+1)2!⋅x2−n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞\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
⇒(1−x)n= 1−nx+n(n+1)2!⋅x2−n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞\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
⇒(1−x)−n= 1+nx+n(n+1)2!⋅x2+n(n+1)(n+2)3!⋅x3+.....+(−1)rn(n+1)(n+2)...(n+r−1)r!⋅xr+.....∞\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 xx will depend on the question.