Solveeit Logo
Login

Question

Question: How do you use the binomial series to expand \({(1 - 4x)^{\dfrac{1}{2}}}\)?...

How do you use the binomial series to expand (1−4x)12{(1 - 4x)^{\dfrac{1}{2}}}?

Explanation

Solution

Use the binomial expansion formula and find the required coefficients and substitute to get the series
First, from the given equation we have to find the value of yyand the value of nn by comparing it with the general expansion of the binomial series
(1+y)n=∑k=0∞(kn)yk{(1 + y)^n} = \sum\limits_{k = 0}^\infty {(\mathop {}\limits_k^n ){y^k}} and after expanding the summation series up to some terms, we get (1+y)n=1+ny+n(n−1)2!y2+n(n−1)(n−2)3!y3+....{(1 + y)^n} = 1 + ny + \dfrac{{n(n - 1)}}{{2!}}{y^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{y^3} + .... after finding the values, we are going to substitute in the expansion form, which will give us the required expansion in the question

Complete step by step answer:
First, we are given a expression
(1−4x)12{(1 - 4x)^{\dfrac{1}{2}}},………………..(1)
The general form of binomial expression is of the form (1+y)n{(1 + y)^n} and its expansion is given below
(1+y)n=∑k=0∞(kn)yk{(1 + y)^n} = \sum\limits_{k = 0}^\infty {(\mathop {}\limits_k^n ){y^k}}
From this summation series, we are going to expand the series upto some terms, we will get the following
(1+y)n=1+ny+n(n−1)2!y2+n(n−1)(n−2)3!y3+....{(1 + y)^n} = 1 + ny + \dfrac{{n(n - 1)}}{{2!}}{y^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{y^3} + ....……………………….(2)
By comparing the given and the LHS of the general expression, we are going to get the value of yyand nn,they are
y = −4xy{\text{ }} = {\text{ }} - 4x
n=12n = \dfrac{1}{2}
Since we have the necessary values for forming the expansion of the binomial series, we are going to substitute them into the equation (2) which is expanded upto some n terms.
(1−4x)12=1+(12)(−4x)+(12)(−12)2!(−4x)2+(12)(−12)(−32)3!(−4x)3+(12)(−12)(−32)(−52)4!(−4x)4{(1 - 4x)^{\dfrac{1}{2}}} = 1 + \left( {\dfrac{1}{2}} \right)( - 4x) + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)}}{{2!}}{( - 4x)^2} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{( - 4x)^3} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{4!}}{( - 4x)^4}
Now, on further simplification of this expansion, we are going to obtain the following
=1−2x−2x2−4x3−10x4+...= 1 - 2x - 2{x^2} - 4{x^3} - 10{x^4} + ...
Here, is the expanded binomial series of (1−4x)12{(1 - 4x)^{\dfrac{1}{2}}}

Note: Here we don’t need to remember the complete expansion of the binomial series rather if we memorize the summation form, then we can easily derive terms according to our needs without any risk of mistake in the expansion series.