The middle term in the expansion of ${{\left( 1-3x+3{{x}^{2}}-{{x}^{3}} \right)}^{2n}}$ is
A.$^{6n}{{C}_{3n}}{{\left( -x \right)}^{3n}}$
B.$^{6n}{{C}_{2n}}{{\left( -x \right)}^{2n+1}}$
C.$^{4n}{{C}_{3n}}{{\left( -x \right)}^{3n}}$
D.$^{6n}{{C}_{3n}}{{\left( -x \right)}^{3n-1}}$
Answer
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Hint: Focus on the point that $1-3x+3{{x}^{2}}-{{x}^{3}}$ can be written as ${{\left( 1-x \right)}^{3}}$ . Also, you should remember that the middle term of the expansion of ${{\left( 1-x \right)}^{2k}}$ is the ${{\left( k+1 \right)}^{th}}$ and is given by $^{2k}{{C}_{k}}{{\left( -x \right)}^{2k-k}}{{=}^{2k}}{{C}_{k}}{{\left( -x \right)}^{k}}$ .
Complete step-by-step answer:
Let us start the solution to the above question by simplifying the expression given in the question. We know that $1-3x+3{{x}^{2}}-{{x}^{3}}$ can be written as ${{\left( 1-x \right)}^{3}}$ . So, our equation becomes:
${{\left( 1-3x+3{{x}^{2}}-{{x}^{3}} \right)}^{2n}}$
\[={{\left( {{\left( 1-x \right)}^{3}} \right)}^{2n}}\]
Now, we know that ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$ . So, if we use this identity, we get
\[{{\left( 1-x \right)}^{3\times 2n}}\]
\[={{\left( 1-x \right)}^{6n}}\]
So, we actually have to find the middle term of the expansion of \[{{\left( 1-x \right)}^{6n}}\] .
We know that the middle term of the expansion of ${{\left( 1-x \right)}^{2k}}$ is the ${{\left( k+1 \right)}^{th}}$ and is given by ${{T}_{k+1}}{{=}^{2k}}{{C}_{k}}{{\left( -x \right)}^{2k-k}}{{=}^{2k}}{{C}_{k}}{{\left( -x \right)}^{k}}$ . So, for the expansion of ${{\left( 1-x \right)}^{6n}}$ , k is equal to 3n. Therefore, the middle term is $^{2k}{{C}_{k}}{{\left( -x \right)}^{k}}{{=}^{6n}}{{C}_{3n}}{{\left( -x \right)}^{3n}}$ .
Hence, the answer to the above question is option (a).
Note: You should know that the number of terms in the expansion of ${{\left( a+b \right)}^{n}}$ in (n+1) and the middle term of the expansion is: ${{T}_{\dfrac{n+2}{2}}}$ , if n is even and if n is odd the middle term is: ${{T}_{\dfrac{n+1}{2}}}\text{ and }{{T}_{\dfrac{n+3}{2}}}$ . You should also know that the ${{\left( k+1 \right)}^{th}}$ term in the expansion of ${{\left( a+b \right)}^{n}}$ is ${{T}_{k+1}}{{=}^{n}}{{C}_{k}}{{a}^{n-k}}{{b}^{k}}$ . It is also important to learn the identities related to exponents and algebraic formulas as they are used very often.
Complete step-by-step answer:
Let us start the solution to the above question by simplifying the expression given in the question. We know that $1-3x+3{{x}^{2}}-{{x}^{3}}$ can be written as ${{\left( 1-x \right)}^{3}}$ . So, our equation becomes:
${{\left( 1-3x+3{{x}^{2}}-{{x}^{3}} \right)}^{2n}}$
\[={{\left( {{\left( 1-x \right)}^{3}} \right)}^{2n}}\]
Now, we know that ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$ . So, if we use this identity, we get
\[{{\left( 1-x \right)}^{3\times 2n}}\]
\[={{\left( 1-x \right)}^{6n}}\]
So, we actually have to find the middle term of the expansion of \[{{\left( 1-x \right)}^{6n}}\] .
We know that the middle term of the expansion of ${{\left( 1-x \right)}^{2k}}$ is the ${{\left( k+1 \right)}^{th}}$ and is given by ${{T}_{k+1}}{{=}^{2k}}{{C}_{k}}{{\left( -x \right)}^{2k-k}}{{=}^{2k}}{{C}_{k}}{{\left( -x \right)}^{k}}$ . So, for the expansion of ${{\left( 1-x \right)}^{6n}}$ , k is equal to 3n. Therefore, the middle term is $^{2k}{{C}_{k}}{{\left( -x \right)}^{k}}{{=}^{6n}}{{C}_{3n}}{{\left( -x \right)}^{3n}}$ .
Hence, the answer to the above question is option (a).
Note: You should know that the number of terms in the expansion of ${{\left( a+b \right)}^{n}}$ in (n+1) and the middle term of the expansion is: ${{T}_{\dfrac{n+2}{2}}}$ , if n is even and if n is odd the middle term is: ${{T}_{\dfrac{n+1}{2}}}\text{ and }{{T}_{\dfrac{n+3}{2}}}$ . You should also know that the ${{\left( k+1 \right)}^{th}}$ term in the expansion of ${{\left( a+b \right)}^{n}}$ is ${{T}_{k+1}}{{=}^{n}}{{C}_{k}}{{a}^{n-k}}{{b}^{k}}$ . It is also important to learn the identities related to exponents and algebraic formulas as they are used very often.
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