Question

How do you use the binomial theorem to expand ${\left( {\dfrac{1}{3}x - 9} \right)^{20}}$ ?

Answer 1

Hint : The given question requires us to find the binomial expansion of the given binomial expression ${\left( {\dfrac{1}{3}x - 9} \right)^{20}}$ . 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 step-by-step answer :
We have to find the binomial expansion of ${\left( {\dfrac{1}{3}x - 9} \right)^{20}}$ .
Now, we will first take LCM of the terms inside the brackets. So, we get,
$\Rightarrow {\left( {\dfrac{1}{3}x - 9} \right)^{20}} = \,{\left( {\dfrac{{x - 27}}{3}} \right)^{20}}$
Now, we take the denominator outside the bracket so as to simplify the expression,
$\Rightarrow \dfrac{1}{{{3^{20}}}}{\left( {x - 27} \right)^{20}}$
Now, we have to find the binomial expansion of ${\left( {x - 27} \right)^{20}}$ .
So, using the binomial theorem, the binomial expansion of ${\left( {x + y} \right)^n}$ is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}}$
So, the binomial expansion of ${\left( {x - 27} \right)^{20}}$ is
$\sum\nolimits_{r = 0}^{20} {\left( {^{20}{C_r}} \right){{\left( x \right)}^{20 - r}}{{\left( { - 27} \right)}^r}}$ .
Now, we have to expand the expression
$\sum\nolimits_{r = 0}^{20} {\left( {^{20}{C_r}} \right){{\left( x \right)}^{20 - r}}{{\left( { - 27} \right)}^r}}$ and we are done with the binomial expansion of ${\left( {x - 27} \right)^{20}}$ .
So, we have,
$= \left( {^{20}{C_0}} \right){\left( x \right)^{20}}{\left( { - 27} \right)^0} + \left( {^{20}{C_1}} \right){\left( x \right)^{19}}{\left( { - 27} \right)^1} + ......\left( {^{20}{C_{19}}} \right){\left( x \right)^1}{\left( { - 27} \right)^{19}} + \left( {^{20}{C_{20}}} \right){\left( x \right)^0}{\left( { - 27} \right)^{20}}$
Now, substituting values of combination formulae and simplifying calculations, we get,
$= \left( 1 \right){\left( x \right)^{20}}{\left( { - 27} \right)^0} + \left( {20} \right){\left( x \right)^{19}}{\left( { - 27} \right)^1} + ......\left( {20} \right){\left( x \right)^1}{\left( { - 27} \right)^{19}} + \left( 1 \right){\left( x \right)^0}{\left( { - 27} \right)^{20}}$
Opening brackets and simplifying further, we get the expression as
$= {x^{20}} - 540{x^{19}} + ......\left( {20} \right){\left( { - 27} \right)^{19}}x + {\left( { - 27} \right)^{20}}$
So, Binomial expansion of ${\left( {x - 27} \right)^{20}}$ is ${x^{20}} - 540{x^{19}} + ......\left( {20} \right){\left( { - 27} \right)^{19}}x + {\left( { - 27} \right)^{20}}$ .
Therefore, ${\left( {\dfrac{1}{3}x - 9} \right)^{20}} = \dfrac{1}{{{3^{20}}}}{\left( {x - 27} \right)^{20}} = \dfrac{1}{{{3^{20}}}}\left[ {{x^{20}} - 540{x^{19}} + ......\left( {20} \right){{\left( { - 27} \right)}^{19}}x + {{\left( { - 27} \right)}^{20}}} \right]$
So, the correct answer is “ $\dfrac{1}{{{3^{20}}}}\left[ {{x^{20}} - 540{x^{19}} + ......\left( {20} \right){{\left( { - 27} \right)}^{19}}x + {{\left( { - 27} \right)}^{20}}} \right]$ ”.

Note : The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. One should compute the values of the combination formula with utmost care as it can involve tedious and cumbersome calculations. There can be various ways of representing the answer that are interconvertible into each other.