Question

The coefficient of ${{x}^{n}}$ in the polynomial $\left( x{{+}^{2n}}{{C}_{0}} \right)\left( x{{+}^{2n}}{{C}_{2}} \right)\left( x{{+}^{2n}}{{C}_{4}} \right).......\left( x{{+}^{2n}}{{C}_{2n}} \right)$
A.${{2}^{2n-1}}$
B.${{2}^{2\left( n-1 \right)}}$
C.${{2}^{n}}$
D.${{2}^{n-1}}$

Answer 1

Hint: First observe the brackets of the given expression and try to find the number of brackets which is the highest power of x or degree of the polynomial. Now, get the coefficient of ${{x}^{n}}$ by expanding it. Use relation
${{\left( 1+x \right)}^{2n}}={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}x+{}^{2n}{{C}_{2}}{{x}^{2}}+.......+{}^{2n}{{C}_{2n}}{{x}^{2n}}$

Complete step-by-step answer:

Given polynomial in the question is
$\left( x+{}^{2n}{{C}_{0}} \right)\left( x+{}^{2n}{{C}_{2}} \right)\left( x+{}^{2n}{{C}_{4}} \right).......\left( x+{}^{2n}{{C}_{2n}} \right)$ ……………………………………(i)
Now, we need to observe the above equation and have to determine the highest power of x i.e. degree of the polynomial after expanding it.
So, first of all, we have to find the number of brackets involved in multiplication which will give the maximum power of x as we need to multiply ‘x’ from all the brackets to get maximum power of ‘x’ or degree of polynomial.
Now, we can observe the sequence ${}^{2n}{{C}_{0}},{}^{2n}{{C}_{2}},{}^{2n}{{C}_{4}},........{}^{2n}{{C}_{2n}}$ . The number of terms of the above sequence will give the number of brackets for multiplication.
So, for getting the number of terms of the above sequence, we need to determine the number of terms in the sequence.
$\left( 0,2,4,...........2n \right)$ …………………………………………(ii)
Here, the above sequence is an A.P. with the common difference of ‘2’ and ‘0’ as the first term.
We know the ${{r}^{th}}$ term of any A.P. can be given as
${{T}_{r}}=a+\left( r-1 \right)d$
Where a = first term
d = common difference
${{T}_{r}}$ is denoting the ${{r}^{th}}$ term of the sequence.
So, let ‘2n’ be the ${{r}^{th}}$ term of the sequence given in equation (ii). Hence, we get
${{T}_{r}}=a+\left( r-1 \right)d$
Put $a=0,d=2,{{T}_{r}}=2n$ from the sequence given in equation (ii)
$2n=0+\left( r-1 \right)2$
$2n=2r-2$
$2n+2=2r$
$r=n+1$
Now, we observe that ‘2n’ is the ${{\left( n+1 \right)}^{th}}$ term of the sequence in the equation (ii), or in other words total number of terms in the sequence is $\left( n+1 \right)$.
Hence, there are $\left( n+1 \right)$ brackets involved in the given polynomial from the question. So, after multiplying the brackets, we will have$\left( n+1 \right)$ as the highest power of x and degree of polynomial as well. So, by taking x from all the brackets we get power of ‘x’ as $\left( n+1 \right)$.
But, we need to determine the coefficient of ${{x}^{n}}$ from the given polynomial. So, we need to take ‘x’ from n brackets and other constant terms from one bracket.
So, we have
$\left( x+{}^{2n}{{C}_{0}} \right)\left( x+{}^{2n}{{C}_{2}} \right)\left( x+{}^{2n}{{C}_{4}} \right)........\left( x+{}^{2n}{{C}_{2n}} \right)$
We know that there are $\left( n+1 \right)$ brackets involved. If we take ${}^{2n}{{C}_{0}}$ from first bracket and ‘x’ from all other n brackets for multiplication, we get value as ${}^{2n}{{C}_{0}}{{x}^{n}}$, similarly we take ${}^{2n}{{C}_{2}}$ from second bracket and ‘x’ from all other brackets, we get ${}^{2n}{{C}_{2}}{{x}^{n}}$, similarly, take ${}^{2n}{{C}_{4}}$ from first bracket and ‘x’ from all other n brackets, we get ${}^{2n}{{C}_{4}}{{x}^{n}}$; Hence, the terms related to ${{x}^{n}}$ can be given after expanding the polynomial is
${}^{2n}{{C}_{0}}{{x}^{n}}+{}^{2n}{{C}_{2}}{{x}^{n}}+{}^{2n}{{C}_{4}}{{x}^{n}}+.......{}^{2n}{{C}_{2n}}{{x}^{n}}$
Hence, coefficient of ${{x}^{n}}$ after expanding the polynomial is
$S={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{4}}+.......{}^{2n}{{C}_{2n}}$ …………………………………….(iii)
We know the expression of ${{\left( 1+x \right)}^{2n}}$ can be given using binomial expansion as
${{\left( 1+x \right)}^{2n}}={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}x+{}^{2n}{{C}_{2}}{{x}^{2}}+......{}^{2n}{{C}_{2n}}{{x}^{2n}}$ ……………………..(iv)
Now, put $x=1$ in the equation (iv). So, we get
${{2}^{2n}}={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}\left( 1 \right)+{}^{2n}{{C}_{2}}{{\left( 1 \right)}^{2}}+......{}^{2n}{{C}_{2n}}{{\left( 1 \right)}^{2n}}$
$\Rightarrow {}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}+......{}^{2n}{{C}_{2n}}={{2}^{2n}}$ …………………………….(v)
Now, put $x=-1$ in the equation (iv), so we get
${{\left( 1-1 \right)}^{2n}}={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}{{\left( -1 \right)}^{1}}+{}^{2n}{{C}_{2}}{{\left( -1 \right)}^{2}}+{}^{2n}{{C}_{3}}{{\left( -1 \right)}^{3}}+......{}^{2n}{{C}_{2n}}{{\left( -1 \right)}^{2n}}$
$\Rightarrow 0={}^{2n}{{C}_{0}}-{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}-{}^{2n}{{C}_{3}}+......{}^{2n}{{C}_{2n}}{{\left( -1 \right)}^{2n}}$
$\Rightarrow {}^{2n}{{C}_{0}}-{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}-{}^{2n}{{C}_{3}}+......{}^{2n}{{C}_{2n}}{{\left( -1 \right)}^{2n}}=0$ ……………………….(vi)
Now, add equations (v) and (vi) to get the values of series ‘S’ given in equation (iii). So, we get
$\left( {}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{3}}+......+{}^{2n}{{C}_{2n}} \right)$
$+\left( {}^{2n}{{C}_{0}}-{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}-{}^{2n}{{C}_{3}}+......+{}^{2n}{{C}_{2n}} \right)={{2}^{2n}}+0$
$\Rightarrow \left( {}^{2n}{{C}_{0}}+{}^{2n}{{C}_{0}} \right)+\left( {}^{2n}{{C}_{1}}-{}^{2n}{{C}_{1}} \right)+\left( {}^{2n}{{C}_{2}}-{}^{2n}{{C}_{2}} \right)+\left( {}^{2n}{{C}_{3}}-{}^{2n}{{C}_{3}} \right)+......\left( {}^{2n}{{C}_{2n}}-{}^{2n}{{C}_{2n}} \right)={{2}^{2n}}$
So, we get
$2\cdot {}^{2n}{{C}_{0}}+2\cdot {}^{2n}{{C}_{2}}+2\cdot {}^{2n}{{C}_{4}}+......2\cdot {}^{2n}{{C}_{2n}}={{2}^{2n}}$
$2\left[ {}^{2n}{{C}_{0}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{4}}+........{}^{2n}{{C}_{2n}} \right]={{2}^{2n}}$
$\Rightarrow S={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{4}}+........{}^{2n}{{C}_{2n}}=\dfrac{{{2}^{2n}}}{2}$
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$; So, $\dfrac{{{2}^{2n}}}{{{2}^{1}}}$ can be replaced as ${{2}^{2n-1}}$. So, we get the coefficient of ${{x}^{n}}$ in the given polynomial :-
${}^{2n}{{C}_{0}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{4}}+........{}^{2n}{{C}_{2n}}={{2}^{2n-1}}$
Hence, option (A) is correct.

Note: Expansion of any polynomial of the form $\left( x+{{a}_{1}} \right)\left( x+{{a}_{2}} \right)\left( x+{{a}_{3}} \right)......\left( x+{{a}_{n}} \right)$ can be given as
${{x}^{n}}+$ (sum of ${{a}_{1}},{{a}_{2}}....{{a}_{n}}$ ) ${{x}^{n-1}}$ + (sum of ${{a}_{1}},{{a}_{2}}....{{a}_{n}}$ taking two at a time ) ${{x}^{n-2}}$ + (sum of ${{a}_{1}},{{a}_{2}}....{{a}_{n}}$ taking three at a time ) ${{x}^{n-3}}$ +…………………….+ (products of ${{a}_{1}},{{a}_{2}}....{{a}_{n}}$)
So, we can directly expand the given polynomial using the above relation as well and get coefficient of ${{x}^{n}}$ in the expression of the problem as sum of ${}^{2n}{{C}_{0}},{}^{2n}{{C}_{2}},{}^{2n}{{C}_{4}},........{}^{2n}{{C}_{2n}}$.
One may use the formulae of ${}^{2n}{{C}_{0}}+{}^{2n}{{C}_{2}}+{}^{2n}{{C}_{4}}+........{}^{2n}{{C}_{2n}}={{2}^{2n}}$ and ${}^{2n}{{C}_{0}}-{}^{2n}{{C}_{1}}+{}^{2n}{{C}_{2}}+........{}^{2n}{{C}_{2n}}=0$ directly as well. These are the direct relations from the binomial expansion of ${{\left( 1+x \right)}^{2n}}$.