
How do you use the binomial series to expand ${{\left( x+3 \right)}^{6}}$?
Answer
561k+ views
Hint: From the question it had been asked to expand ${{\left( x+3 \right)}^{6}}$ using the binomial series. To expand the given question, we have to know the formula of general expansion of the type. The above given question ${{\left( x+3 \right)}^{6}}$ can be expanded by using a simple basic expansion formula of binomial series.
Complete step by step answer:
We can clearly observe from the question that it is in the form of ${{\left( x+a \right)}^{n}}$
By comparing the both expressions, we get the values as,
$\begin{align}
& a=3 \\
& n=6 \\
\end{align}$
As we have been already discussed above, to expand the given question first we have to write the general expansion of ${{\left( x+a \right)}^{n}}$
By using the general expansion of ${{\left( x+a \right)}^{n}}$, we can expand the given question very easily.
General expansion of ${{\left( x+a \right)}^{n}}$ is written below,
Expansion of ${{\left( x+a \right)}^{n}}$:
${{\left( x+a \right)}^{n}}={{n}_{{{c}_{0}}}}{{x}^{n}}{{a}^{0}}+{{n}_{{{c}_{1}}}}{{x}^{n-1}}{{a}^{1}}+...............+{{n}_{{{c}_{n}}}}{{x}^{0}}{{a}^{n}}$
By using the above expansion formula, we can get the expansion of ${{\left( x+3 \right)}^{6}}$
By applying the above formula for expansion, we get
By substituting the value of $n=6$ in the expansion we get,
${{\left( x+a \right)}^{6}}={{x}^{6}}+6{{x}^{5}}a+15{{x}^{4}}{{a}^{2}}+20{{x}^{3}}{{a}^{3}}+15{{x}^{2}}{{a}^{4}}+6x{{a}^{5}}+{{a}^{6}}$
Now to further more simplify the given expansion we have to substitute $a=3$ in the above expansion.
By substituting $a=3$ in the above expansion we get,
${{\left( x+3 \right)}^{6}}={{x}^{6}}+6{{x}^{5}}\left( 3 \right)+15{{x}^{4}}{{\left( 3 \right)}^{2}}+20{{x}^{3}}{{\left( 3 \right)}^{3}}+15{{x}^{2}}{{\left( 3 \right)}^{4}}+6x{{\left( 3 \right)}^{5}}+{{3}^{6}}$
On furthermore simplifying the expansion we get,
${{\left( x+3 \right)}^{6}}={{x}^{6}}+18{{x}^{5}}+135{{x}^{4}}+540{{x}^{3}}+1215{{x}^{2}}+1458x+729$
Note:
We should be well aware of the binomial series and we should be well known about the expansions of the binomial series. In this type of questions, we should observe that the given question is in which type and which expansion formula is suitable for it. This question can be answered in other ways also but it is not preferable because it will take a lot of our time and it also does not involve binomial series concept. The procedure of solving in other way is ${{\left( x+3 \right)}^{6}}={{\left( {{\left( x+3 \right)}^{2}} \right)}^{3}}={{\left( {{x}^{2}}+9+6x \right)}^{3}}={{x}^{6}}+18{{x}^{5}}+135{{x}^{4}}+540{{x}^{3}}+1215{{x}^{2}}+1458x+729$ .
Complete step by step answer:
We can clearly observe from the question that it is in the form of ${{\left( x+a \right)}^{n}}$
By comparing the both expressions, we get the values as,
$\begin{align}
& a=3 \\
& n=6 \\
\end{align}$
As we have been already discussed above, to expand the given question first we have to write the general expansion of ${{\left( x+a \right)}^{n}}$
By using the general expansion of ${{\left( x+a \right)}^{n}}$, we can expand the given question very easily.
General expansion of ${{\left( x+a \right)}^{n}}$ is written below,
Expansion of ${{\left( x+a \right)}^{n}}$:
${{\left( x+a \right)}^{n}}={{n}_{{{c}_{0}}}}{{x}^{n}}{{a}^{0}}+{{n}_{{{c}_{1}}}}{{x}^{n-1}}{{a}^{1}}+...............+{{n}_{{{c}_{n}}}}{{x}^{0}}{{a}^{n}}$
By using the above expansion formula, we can get the expansion of ${{\left( x+3 \right)}^{6}}$
By applying the above formula for expansion, we get
By substituting the value of $n=6$ in the expansion we get,
${{\left( x+a \right)}^{6}}={{x}^{6}}+6{{x}^{5}}a+15{{x}^{4}}{{a}^{2}}+20{{x}^{3}}{{a}^{3}}+15{{x}^{2}}{{a}^{4}}+6x{{a}^{5}}+{{a}^{6}}$
Now to further more simplify the given expansion we have to substitute $a=3$ in the above expansion.
By substituting $a=3$ in the above expansion we get,
${{\left( x+3 \right)}^{6}}={{x}^{6}}+6{{x}^{5}}\left( 3 \right)+15{{x}^{4}}{{\left( 3 \right)}^{2}}+20{{x}^{3}}{{\left( 3 \right)}^{3}}+15{{x}^{2}}{{\left( 3 \right)}^{4}}+6x{{\left( 3 \right)}^{5}}+{{3}^{6}}$
On furthermore simplifying the expansion we get,
${{\left( x+3 \right)}^{6}}={{x}^{6}}+18{{x}^{5}}+135{{x}^{4}}+540{{x}^{3}}+1215{{x}^{2}}+1458x+729$
Note:
We should be well aware of the binomial series and we should be well known about the expansions of the binomial series. In this type of questions, we should observe that the given question is in which type and which expansion formula is suitable for it. This question can be answered in other ways also but it is not preferable because it will take a lot of our time and it also does not involve binomial series concept. The procedure of solving in other way is ${{\left( x+3 \right)}^{6}}={{\left( {{\left( x+3 \right)}^{2}} \right)}^{3}}={{\left( {{x}^{2}}+9+6x \right)}^{3}}={{x}^{6}}+18{{x}^{5}}+135{{x}^{4}}+540{{x}^{3}}+1215{{x}^{2}}+1458x+729$ .
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