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# How do you use the binomial theorem to expand ${{\left( 2x-1 \right)}^{4}}$?

Last updated date: 21st Jun 2024
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Hint:Binomial theorem is a method used to expand a binomial term that is raised to some power of positive integer. According to binomial theorem, the nth power of the sum of two numbers (say a and b) can be expressed (expanded) as the sum or series of (n+1) terms, provided that ‘n’ is a positive integer.

Formula used:
${{(x+y)}^{n}}=\sum\limits_{i=0}^{n}{{}^{n}{{C}_{i}}{{x}^{n-i}}{{y}^{i}}}$,
where x and y are real numbers and n is a positive integer (a natural number).
${}^{n}{{C}_{i}}=\dfrac{n!}{i!(n-i)!}$

Let us first understand what is the binomial theorem.Binomial theorem is a method used to expand a binomial term that is raised to some power of positive integer. According to binomial theorem, the nth power of the sum of two numbers (say a and b) can be expressed (expanded) as the sum or series of (n+1) terms, provided that ‘n’ is a positive integer.
Suppose we have an expression ${{(x+y)}^{n}}$, where x and y are real numbers and n is a positive integer (a natural number).

Then, the binomial expansion of the above expression is given as
${{(x+y)}^{n}}=\sum\limits_{i=0}^{n}{{}^{n}{{C}_{i}}{{x}^{n-i}}{{y}^{i}}}$
Here, i is a natural number taking values from 0 to n.
When we expand the summation we get that ${{(x+y)}^{n}}={}^{n}{{C}_{0}}{{x}^{n-0}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+.......+{}^{n}{{C}_{n-1}}{{x}^{n-(n-1)}}{{y}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n-n}}{{y}^{n}}$.
In the given question, $n=4$,
Therefore, the given expression can expanded, with the help of binomial theorem as
${{(2x-1)}^{4}}={}^{4}{{C}_{0}}{{(2x)}^{4-0}}{{(-1)}^{0}}+{}^{4}{{C}_{1}}{{(2x)}^{4-1}}{{(-1)}^{1}}+{}^{4}{{C}_{2}}{{(2x)}^{4-2}}{{(-1)}^{2}}+{}^{4}{{C}_{3}}{{(2x)}^{4-3}}{{(-1)}^{3}}+{}^{4}{{C}_{4}}{{(2x)}^{4-4}}{{(-1)}^{4}}$
This equation can be further simplified to
${{(2x-1)}^{4}}={}^{4}{{C}_{0}}{{(2x)}^{4}}{{(-1)}^{0}}+{}^{4}{{C}_{1}}{{(2x)}^{3}}{{(-1)}^{1}}+{}^{4}{{C}_{2}}{{(2x)}^{2}}{{(-1)}^{2}}+{}^{4}{{C}_{3}}{{(2x)}^{1}}{{(-1)}^{3}}+{}^{4}{{C}_{4}}{{(2x)}^{0}}{{(-1)}^{4}}$
$\Rightarrow {{(2x-1)}^{4}}={}^{4}{{C}_{0}}(16{{x}^{4}})-{}^{4}{{C}_{1}}(8{{x}^{3}})+{}^{4}{{C}_{2}}(4{{x}^{2}})-{}^{4}{{C}_{3}}(2x)+{}^{4}{{C}_{4}}(1)$ ….. (i)
Now, we shall use the formula ${}^{n}{{C}_{i}}=\dfrac{n!}{i!(n-i)!}$

Therefore, equation (i) can be simplified to
${{(2x-1)}^{4}}=\dfrac{4!}{0!(4-0)!}(16{{x}^{4}})-\dfrac{4!}{1!(4-1)!}(8{{x}^{3}})+\dfrac{4!}{2!(4-2)!}(4{{x}^{2}})-\dfrac{4!}{3!(4-3)!}(2x)+\dfrac{4!}{4!(4-4)!}(1)$
With this, we get that
${{(2x-1)}^{4}}=(1)(16{{x}^{4}})-\dfrac{4!}{1!3!}(8{{x}^{3}})+\dfrac{4!}{2!2!}(4{{x}^{2}})-\dfrac{4!}{3!1!}(2x)+\dfrac{4!}{4!0!}(1)$
$\Rightarrow {{(2x-1)}^{4}}=16{{x}^{4}}-(4)8{{x}^{3}}+\left( \dfrac{4\times 3}{2} \right)(4{{x}^{2}})-(4)(2x)+(1)$
Finally,
$\therefore {{(2x-1)}^{4}}=16{{x}^{4}}-32{{x}^{3}}+24{{x}^{2}}-8x+1$
Hence, we found the expansion of the given expression with the help of binomial theorem.

Note:when we expand an expression with the help of binomial theorem, the series consists of (n+1) terms. If you do not use the formula of combination ${}^{n}{{C}_{i}}$, then you can make use of Pascal's triangle and select the row that has (n+1) elements (numbers).