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# How do you expand ${{\left( 2{{x}^{3}}+1 \right)}^{5}}$?

Last updated date: 20th Jun 2024
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Hint: Binomial theorem is a method used to expand a binomial term that is raise 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).
${}^{n}{{C}_{i}}=\dfrac{n!}{i!(n-i)!}$

Complete step-by-step solution:
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, $n=5$,
Therefore, the given expression can expanded, with the help of binomial theorem as
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}={}^{5}{{C}_{0}}{{(2{{x}^{3}})}^{5-0}}{{(1)}^{0}}+{}^{5}{{C}_{1}}{{(2{{x}^{3}})}^{5-1}}{{(1)}^{1}}+{}^{5}{{C}_{2}}{{(2{{x}^{3}})}^{5-2}}{{(1)}^{2}}+{}^{5}{{C}_{3}}{{(2{{x}^{3}})}^{5-3}}{{(1)}^{3}}+{}^{5}{{C}_{4}}{{(2{{x}^{3}})}^{5-4}}{{(1)}^{4}}+{}^{5}{{C}_{5}}{{(2{{x}^{3}})}^{5-5}}{{(1)}^{5}}$
This equation can be further simplified to
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}={}^{5}{{C}_{0}}{{(2{{x}^{3}})}^{5}}+{}^{5}{{C}_{1}}{{(2{{x}^{3}})}^{4}}+{}^{5}{{C}_{2}}{{(2{{x}^{3}})}^{3}}+{}^{5}{{C}_{3}}{{(2{{x}^{3}})}^{2}}+{}^{5}{{C}_{4}}{{(2{{x}^{3}})}^{1}}+{}^{5}{{C}_{5}}{{(2{{x}^{3}})}^{0}}$
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}={}^{5}{{C}_{0}}(32{{x}^{15}})+{}^{5}{{C}_{1}}(16{{x}^{12}})+{}^{5}{{C}_{2}}(8{{x}^{9}})+{}^{5}{{C}_{3}}(4{{x}^{6}})+{}^{5}{{C}_{4}}(2{{x}^{3}})+{}^{5}{{C}_{5}}(1)$ ….. (i)
Now, we shall use the formula ${}^{n}{{C}_{i}}=\dfrac{n!}{i!(n-i)!}$
Therefore, equation (i) can be simplified to
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}=\dfrac{5!}{0!(5-0)!}(32{{x}^{15}})-\dfrac{5!}{1!(5-1)!}(16{{x}^{12}})+\dfrac{5!}{2!(5-2)!}(8{{x}^{9}})-\dfrac{5!}{3!(5-3)!}(4{{x}^{6}})+\dfrac{5!}{4!(5-4)!}(2{{x}^{3}})+\dfrac{5!}{5!(5-5)!}$
With this, we get that
$\Rightarrow{{(2{{x}^{3}}+1)}^{5}}=\dfrac{5!}{5!}(32{{x}^{15}})+\dfrac{5!}{1!(4)!}(16{{x}^{12}})+\dfrac{5!}{2!(3)!}(8{{x}^{9}})+\dfrac{5!}{3!(2)!}(4{{x}^{6}})+\dfrac{5!}{4!(1)!}(2{{x}^{3}})+\dfrac{5!}{5!(0)!}$
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}=32{{x}^{15}}-(5)(16{{x}^{12}})+\left( \dfrac{5\times 4}{2} \right)(8{{x}^{9}})+\left( \dfrac{5\times 4\times 3}{3\times 2} \right)(4{{x}^{6}})+(5)(2{{x}^{3}})+1$
Finally,
$\Rightarrow {{(2{{x}^{3}}+1)}^{5}}=32{{x}^{15}}+80{{x}^{12}})+80{{x}^{9}}+40{{x}^{6}}+10{{x}^{3}})+1$
Hence, we found the expansion of the given expression with the help of binomial theorem.

Note: Note that 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).