Question

What is the binomial expansion of ${{\left( 2x-1 \right)}^{5}}$ ?

Answer 1

In this question, we will first write our expression as, ${{\left( -1 \right)}^{5}}{{\left( 1-2x \right)}^{5}}$ . Now, the second term in our resulting expression could be expanded with the help of standard binomial expansion formula. This formula is given can be written as:
$\Rightarrow {{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+.......$ . We shall proceed with our solution in this manner.

Complete step by step answer:
We have been given the mathematical expression in our problem as: ${{\left( 2x-1 \right)}^{5}}$
Taking ${{\left( -1 \right)}^{5}}$ common from our expression, we can write the new expression as:
$\begin{aligned} & \Rightarrow {{\left( 2x-1 \right)}^{5}}={{\left( -1 \right)}^{5}}{{\left( 1-2x \right)}^{5}} \\\ & \therefore {{\left( 2x-1 \right)}^{5}}=-{{\left( 1-2x \right)}^{5}} \\\ \end{aligned}$
Now, in the binomial expansion formula, we can replace the term $\left( x \right)$ by the term $\left( -2x \right)$ to get our required solution. While writing the expansion series, we will include terms only up to the term with the highest degree in $\left( x \right)$ . The highest degree term in our expansion will be ${{x}^{5}}$, so we will include the term containing ${{x}^{5}}$ as the last term of our series and no further terms will be included.
Thus, we have:
$\Rightarrow {{\left( 2x-1 \right)}^{5}}=-{{\left[ 1+\left( -2x \right) \right]}^{5}}$
Applying the binomial expansion formula in the Right-Hand Side of our equation, we get:$\Rightarrow {{\left( 2x-1 \right)}^{5}}=-\left[ \begin{aligned} & 1+5\left( -2x \right)+\dfrac{5\left( 5-1 \right)}{2!}{{\left( -2x \right)}^{2}}+\dfrac{5\left( 5-1 \right)\left( 5-2 \right)}{3!}{{\left( -2x \right)}^{3}}+\dfrac{5\left( 5-1 \right)\left( 5-2 \right)\left( 5-3 \right)}{4!}{{\left( -2x \right)}^{4}} \\\ & +\dfrac{5\left( 5-1 \right)\left( 5-2 \right)\left( 5-3 \right)\left( 5-4 \right)}{5!}{{\left( -2x \right)}^{5}} \\\ \end{aligned} \right]$
Simplifying all the terms in our equation, we get the new expression as:
$\Rightarrow {{\left( 2x-1 \right)}^{5}}=-\left[ 1-10x+\dfrac{5\times 4}{2\times 1}\left( 4{{x}^{2}} \right)-\dfrac{5\times 4\times 3}{3\times 2\times 1}\left( 8{{x}^{3}} \right)+\dfrac{5\times 4\times 3\times 2}{4\times 3\times 2\times 1}\left( 16{{x}^{4}} \right)-\dfrac{5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}\left( 32{{x}^{5}} \right) \right]$
On further simplification, we get the end result of our calculation as:
$\begin{aligned} & \Rightarrow {{\left( 2x-1 \right)}^{5}}=-\left[ 1-10x+40{{x}^{2}}-80{{x}^{3}}+80{{x}^{4}}-32{{x}^{5}} \right] \\\ & \therefore {{\left( 2x-1 \right)}^{5}}=-1+10x-40{{x}^{2}}+80{{x}^{3}}-80{{x}^{4}}+32{{x}^{5}} \\\ \end{aligned}$

Hence, the binomial expansion of ${{\left( 2x-1 \right)}^{5}}$ comes out to be $-1+10x-40{{x}^{2}}+80{{x}^{3}}-80{{x}^{4}}+32{{x}^{5}}$

Note: The binomial expansion is a very useful and important formula for expanding any mathematical expression. The standard formulas of ${{\left( a+b \right)}^{2}}$ and ${{\left( a+b \right)}^{3}}$ have all been derived with the help of this binomial expansion. All the terms in a binomial expansion of ${{\left( a+b \right)}^{m}}$ have their coefficient as ${{a}^{i}}{{b}^{j}}$ which can be used as a tool to validate our binomial expansion in problems.