Expand the following binomial:
${{(2x-y)}^{5}}$ .
Answer
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Hint: We have to expand ${{(2x-y)}^{5}}$ , for that use formula \[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\] and assume $a=2x$, $b=-y$ and $n=5$ . Try it, you will get the answer.
Complete step-by-step answer:
As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial Theorem. Learn about all the details about binomial theorem like its definition, properties, applications.
According to the binomial theorem, the ${{(r+1)}^{th}}$term in the expansion of ${{(a+b)}^{n}}$is,
\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
The above term is a general term or${{(r+1)}^{th}}$term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$is$(n+1)$, i.e. one more than the exponent$n$.
In the Binomial expression, we have
\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
So the coefficients${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.
You can see them${}^{n}{{C}_{r}}$being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$because, as we know that,
$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$
Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to${{2}^{n-1}}$.
The middle term depends upon the value of$n$,
If $n$ is even: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (odd).
If $n$ is odd: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (even).
If $n$is a positive integer,
\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
So here $a=2x$, $b=-y$ and $n=5$ .
So using the binomial expansion,
\[{{(2x-y)}^{5}}={}^{5}{{C}_{0}}{{(2x)}^{5}}{{\left( -y \right)}^{0}}+{}^{5}{{C}_{1}}{{(2x)}^{5-1}}{{\left( -y \right)}^{1}}+{}^{5}{{C}_{2}}{{(2x)}^{5-2}}{{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}{{(2x)}^{5-3}}{{\left( -y \right)}^{3}}+{}^{4}{{C}_{4}}{{(2x)}^{5-4}}{{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{(2x)}^{5-5}}{{\left( -y \right)}^{5}}\]
\[\begin{align}
& {{(2x-y)}^{5}}={}^{5}{{C}_{0}}{{(2x)}^{5}}+{}^{5}{{C}_{1}}{{(2x)}^{4}}\left( -y \right)+{}^{5}{{C}_{2}}{{(2x)}^{3}}{{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}{{(2x)}^{2}}{{\left( -y \right)}^{3}}+{}^{5}{{C}_{4}}{{(2x)}^{1}}{{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{(2x)}^{0}}{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})+{}^{5}{{C}_{1}}(16{{x}^{4}})\left( -y \right)+{}^{5}{{C}_{2}}(8{{x}^{3}}){{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}(4{{x}^{2}}){{\left( -y \right)}^{3}}+{}^{5}{{C}_{4}}(2x){{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})+5(16{{x}^{4}})\left( -y \right)+\dfrac{5\times 4}{2\times 1}(8{{x}^{3}}){{\left( -y \right)}^{2}}+\dfrac{5\times 4\times 3}{3\times 2\times 1}(4{{x}^{2}}){{\left( -y \right)}^{3}}+\dfrac{5\times 4\times 3\times 2}{4\times 3\times 2\times 1}(2x){{\left( -y \right)}^{4}}+{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})-80{{x}^{4}}y+80{{x}^{3}}{{y}^{2}}-40{{x}^{2}}{{y}^{3}}+10x{{y}^{4}}-{{y}^{5}} \\
\end{align}\]
So we get the expansion as,
\[{{(2x-y)}^{5}}=(32{{x}^{5}})-80{{x}^{4}}y+80{{x}^{3}}{{y}^{2}}-40{{x}^{2}}{{y}^{3}}+10x{{y}^{4}}-{{y}^{5}}\] .
Note: Read the question and see what is asked. Your concept regarding binomial theorem should be clear. Also, you must know the general formula of the binomial theorem. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.
Complete step-by-step answer:
As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial Theorem. Learn about all the details about binomial theorem like its definition, properties, applications.
According to the binomial theorem, the ${{(r+1)}^{th}}$term in the expansion of ${{(a+b)}^{n}}$is,
\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
The above term is a general term or${{(r+1)}^{th}}$term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$is$(n+1)$, i.e. one more than the exponent$n$.
In the Binomial expression, we have
\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
So the coefficients${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.
You can see them${}^{n}{{C}_{r}}$being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$because, as we know that,
$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$
Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to${{2}^{n-1}}$.
The middle term depends upon the value of$n$,
If $n$ is even: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (odd).
If $n$ is odd: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (even).
If $n$is a positive integer,
\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
So here $a=2x$, $b=-y$ and $n=5$ .
So using the binomial expansion,
\[{{(2x-y)}^{5}}={}^{5}{{C}_{0}}{{(2x)}^{5}}{{\left( -y \right)}^{0}}+{}^{5}{{C}_{1}}{{(2x)}^{5-1}}{{\left( -y \right)}^{1}}+{}^{5}{{C}_{2}}{{(2x)}^{5-2}}{{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}{{(2x)}^{5-3}}{{\left( -y \right)}^{3}}+{}^{4}{{C}_{4}}{{(2x)}^{5-4}}{{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{(2x)}^{5-5}}{{\left( -y \right)}^{5}}\]
\[\begin{align}
& {{(2x-y)}^{5}}={}^{5}{{C}_{0}}{{(2x)}^{5}}+{}^{5}{{C}_{1}}{{(2x)}^{4}}\left( -y \right)+{}^{5}{{C}_{2}}{{(2x)}^{3}}{{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}{{(2x)}^{2}}{{\left( -y \right)}^{3}}+{}^{5}{{C}_{4}}{{(2x)}^{1}}{{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{(2x)}^{0}}{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})+{}^{5}{{C}_{1}}(16{{x}^{4}})\left( -y \right)+{}^{5}{{C}_{2}}(8{{x}^{3}}){{\left( -y \right)}^{2}}+{}^{5}{{C}_{3}}(4{{x}^{2}}){{\left( -y \right)}^{3}}+{}^{5}{{C}_{4}}(2x){{\left( -y \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})+5(16{{x}^{4}})\left( -y \right)+\dfrac{5\times 4}{2\times 1}(8{{x}^{3}}){{\left( -y \right)}^{2}}+\dfrac{5\times 4\times 3}{3\times 2\times 1}(4{{x}^{2}}){{\left( -y \right)}^{3}}+\dfrac{5\times 4\times 3\times 2}{4\times 3\times 2\times 1}(2x){{\left( -y \right)}^{4}}+{{\left( -y \right)}^{5}} \\
& {{(2x-y)}^{5}}=(32{{x}^{5}})-80{{x}^{4}}y+80{{x}^{3}}{{y}^{2}}-40{{x}^{2}}{{y}^{3}}+10x{{y}^{4}}-{{y}^{5}} \\
\end{align}\]
So we get the expansion as,
\[{{(2x-y)}^{5}}=(32{{x}^{5}})-80{{x}^{4}}y+80{{x}^{3}}{{y}^{2}}-40{{x}^{2}}{{y}^{3}}+10x{{y}^{4}}-{{y}^{5}}\] .
Note: Read the question and see what is asked. Your concept regarding binomial theorem should be clear. Also, you must know the general formula of the binomial theorem. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.
Last updated date: 21st Sep 2023
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