Question

How do you use the binomial series to expand $ {\left( {x - 2} \right)^5} $ ?

Answer 1

Hint: In order to determine the expansion of the above function using binomial theorem, compare it with the form $ {\left( {x + y} \right)^n} $ to find out the value of $ x $ and $ y $ and then use the binomial theorem $ {\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $ by putting the value of $ x $ and $ y $ and expand the summation. Simplifying the equation, you will get your required result.

Complete step-by-step answer:
(i)
We are given a function $ {\left( {x - 2} \right)^5} $ , and we have to expand this function with the help of binomial theorem.
Let’s first understand why we are using binomial theorem. The reason is pretty simple, because the power of the function is large. When the power of any function increases, the expansion becomes really very tough and lengthy.
In such situations, binomial theorems come into play which provide a simple formula to expand any function of the form $ {\left( {x + y} \right)^n} $ with the use of $ ^n{C_r} $ .
According to the Binomial Theorem:
 $ {\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $
Where, $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ , $ n \in N,x,y \in R $
If we expand the summation, we get
 $ {\left( {x + y} \right)^n}{ = ^n}{C_0}{x^n}{y^0}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2}{ + ^n}{C_3}{x^{n - 3}}{y^3} + .....{ + ^n}{C_n}{x^0}{y^n} $
According to our question, we have to expand $ {\left( {x - 2} \right)^5} $ , so if we compare this with $ {\left( {x + y} \right)^n} $
We get $ n = 5 $ , $ x = x $ and $ y = - 2 $
(ii)
Now, putting these values in the binomial theorem, we get:
\[{\left( {x - 2} \right)^5} = \sum\limits_{r = 0}^5 {^5{C_r}{{\left( x \right)}^{5 - r}}{{\left( { - 2} \right)}^r}} \]
Expanding the summation, we get:
\[{\left( {x - 2} \right)^5}{ = ^5}{C_0}{\left( x \right)^{5 - 0}}{\left( { - 2} \right)^0}{ + ^5}{C_1}{\left( x \right)^{5 - 1}}{\left( { - 2} \right)^1}{ + ^5}{C_2}{\left( x \right)^{5 - 2}}{\left( { - 2} \right)^2}{ + ^5}{C_3}{\left( x \right)^{5 - 3}}{\left( { - 2} \right)^3}{ + ^5}{C_4}{\left( x \right)^{5 - 4}}{\left( { - 2} \right)^4}{ + ^5}{C_5}{\left( x \right)^{5 - 5}}{\left( { - 2} \right)^5}\]
(iii)
Simplifying it further, we get:
\[
  {\left( {x - 2} \right)^5}{ = ^5}{C_0}{\left( x \right)^5}{\left( { - 2} \right)^0}{ + ^5}{C_1}{\left( x \right)^4}{\left( { - 2} \right)^1}{ + ^5}{C_2}{\left( x \right)^3}{\left( { - 2} \right)^2}{ + ^5}{C_3}{\left( x \right)^2}{\left( { - 2} \right)^3}{ + ^5}{C_4}{\left( x \right)^1}{\left( { - 2} \right)^4}{ + ^5}{C_5}{\left( x \right)^0}{\left( { - 2} \right)^5} \\
  {\left( {x - 2} \right)^5}{ = ^5}{C_0}\left( {{x^5}} \right)\left( 1 \right){ + ^5}{C_1}\left( {{x^4}} \right)\left( { - 2} \right){ + ^5}{C_2}\left( {{x^3}} \right)\left( 4 \right){ + ^5}{C_3}\left( {{x^2}} \right)\left( { - 8} \right){ + ^5}{C_4}\left( x \right)\left( {16} \right){ + ^5}{C_5}\left( 1 \right)\left( { - 32} \right) \;
 \]
(iv)
Now, as we know that:
\[
  ^5{C_0} = \dfrac{{5!}}{{0!5!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{1 \times 5 \times 4 \times 3 \times 2 \times 1}} = 1 \\
  ^5{C_1} = \dfrac{{5!}}{{1!4!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{1 \times 4 \times 3 \times 2 \times 1}} = 5 \\
  ^5{C_2} = \dfrac{{5!}}{{2!3!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} = 10 \\
  ^5{C_3} = \dfrac{{5!}}{{3!2!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}} = 10 \\
  ^5{C_4} = \dfrac{{5!}}{{4!1!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 1}} = 5 \\
  ^5{C_5} = \dfrac{{5!}}{{5!0!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1 \times 1}} = 1 \;
 \]
Therefore, putting these values in our obtained equation, we will get:
\[
  {\left( {x - 2} \right)^5} = \left( 1 \right)\left( {{x^5}} \right)\left( 1 \right) + \left( 5 \right)\left( {{x^4}} \right)\left( { - 2} \right) + \left( {10} \right)\left( {{x^3}} \right)\left( 4 \right) + \left( {10} \right)\left( {{x^2}} \right)\left( { - 8} \right) + \left( 5 \right)\left( x \right)\left( {16} \right) + \left( 1 \right)\left( 1 \right)\left( { - 32} \right) \\
  {\left( {x - 2} \right)^5} = {x^5} - 10{x^4} + 40{x^3} - 80{x^2} + 80x - 32 \\
 \]
Hence, the binomial expansion of the function $ {\left( {x - 2} \right)^5} $ is
\[{\left( {x - 2} \right)^5} = {x^5} - 10{x^4} + 40{x^3} - 80{x^2} + 80x - 32\]
So, the correct answer is “\[{\left( {x - 2} \right)^5} = {x^5} - 10{x^4} + 40{x^3} - 80{x^2} + 80x - 32\]”.

Note: An important thing to note is that the powers of each term keep decreasing until it reaches the last term. The pattern of the power is written in such a way that two variables of power addition should be a sum of $ 5 $ . For example, $ 0 + 5 = 5 $ , $ 1 + 4 = 5 $ , $ 2 + 3 = 5 $ , $ 3 + 2 = 5 $ , $ 4 + 1 = 5 $ and $ 5 + 0 = 5 $ .