Question

Prove that coefficient of in ${{x}^{n}}$ the expansion of ${{\left( 1+x \right)}^{2n}}$ is twice the coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n-1}}$.
(a) True.
(b) False.

Answer 1

Hint: We will use binomial theorem to write expanded form of ${{\left( 1+x \right)}^{2n}}$ and ${{\left( 1+x \right)}^{2n-1}}$. Then we will find the coefficient of ${{x}^{n}}$ in both expansions and check if the condition given is true or false.
Complete step by step answer:
Firstly, we will use binomial theorem to write the expansion of ${{\left( 1+x \right)}^{2n}}$ and ${{\left( 1+x \right)}^{2n-1}}$.
Binomial theorem states that,
For any positive integer $n$, the ${{n}^{th}}$ power of the sum of two real numbers $a$ and $b$ may be expressed as the sum of $n+1$ terms as given below:
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$
 $\Rightarrow {{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+\cdots +{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+\cdots +{}^{n}{{C}_{n}}{{b}^{n}}$.
Where, ${}^{n}{{C}_{r}}=\dfrac{\left| \!{\underline {\,
  n \,}} \right. }{\left| \!{\underline {\,
  r \,}} \right. \times \left| \!{\underline {\,
  n-r \,}} \right. }$.
Now, to write expansion of ${{\left( 1+x \right)}^{2n}}$, we compare ${{\left( 1+x \right)}^{2n}}$ with ${{\left( a+b \right)}^{n}},$ so we get,
$a=1,b=x,n=2n$.
Hence, by putting these values in the binomial theorem, we can write,
$\Rightarrow {{\left( 1+x \right)}^{2n}}={}^{2n}{{C}_{0}}{{1}^{2n}}+{}^{2n}{{C}_{1}}{{1}^{2n-1}}{{x}^{1}}+{}^{2n}{{C}_{2}}{{1}^{2n-2}}{{x}^{2}}+\cdots +{}^{2n}{{C}_{n}}{{1}^{2n-n}}{{x}^{n}}+\cdots +{}^{2n}{{C}_{2n}}{{x}^{2n}}$
$={}^{2n}{{C}_{0}}+{}^{2n}{{C}_{1}}{{x}^{1}}+{}^{2n}{{C}_{2}}{{x}^{2}}+\cdots +{}^{2n}{{C}_{n}}{{x}^{n}}+\cdots +{}^{2n}{{C}_{2n}}{{x}^{2n}}$.
In this expansion, the coefficient of ${{x}^{n}}$ is ${}^{2n}{{C}_{n}}$.
Here, ${}^{2n}{{C}_{n}}=\dfrac{\left| \!{\underline {\,
  2n \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  2n-n \,}} \right. }$
$=\dfrac{\left| \!{\underline {\,
  2n \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  n \,}} \right. }$
From definition of factorial, we can write,
$=\dfrac{2n\times \left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times n\times \left| \!{\underline {\,
  n-1 \,}} \right. }$
Cancelling $n$, we get,
$=\dfrac{2\left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  nF-1 \,}} \right. }$
$=2\times \left( \dfrac{\left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  n-1 \,}} \right. } \right)\cdots \cdots \left( i \right)$
Again, to write expansion of ${{\left( 1+x \right)}^{2n-1}}$, we compare ${{\left( 1+x \right)}^{2n-1}}$ with ${{\left( a+b \right)}^{n}},$ so we get,
$a=1,b=x,n=2n-1$.
Hence, by putting these values in the binomial theorem, we can write,
$\Rightarrow {{\left( 1+x \right)}^{2n-1}}={}^{2n-1}{{C}_{0}}{{1}^{2n-1}}+{}^{2n-1}{{C}_{1}}{{1}^{2n-1-1}}{{x}^{1}}+{}^{2n-1}{{C}_{2}}{{1}^{2n-1-2}}{{x}^{2}}+\cdots +{}^{2n-1}{{C}_{n}}{{1}^{2n-1-n}}{{x}^{n}}+\cdots +{}^{2n-1}{{C}_{2n-1}}{{x}^{2n-1}}$
$={}^{2n-1}{{C}_{0}}+{}^{2n-1}{{C}_{1}}{{x}^{1}}+{}^{2n-1}{{C}_{2}}{{x}^{2}}+\cdots +{}^{2n-1}{{C}_{n}}{{x}^{n}}+\cdots +{}^{2n-1}{{C}_{2n-1}}{{x}^{2n-1}}$.
In this expansion, the coefficient of ${{x}^{n}}$ is ${}^{2n-1}{{C}_{n}}$.
Here, ${}^{2n-1}{{C}_{n}}=\dfrac{\left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  2n-1-n \,}} \right. }$
$=\dfrac{\left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  n-1 \,}} \right. }\cdots \cdots \left( ii \right)$
Form, equation $\left( i \right)$ and $\left( ii \right)$, we get,
${}^{2n}{{C}_{n}}=2\times \left( \dfrac{\left| \!{\underline {\,
  2n-1 \,}} \right. }{\left| \!{\underline {\,
  n \,}} \right. \times \left| \!{\underline {\,
  n-1 \,}} \right. } \right)$
$\Rightarrow {}^{2n}{{C}_{n}}=2\times {}^{2n-1}{{C}_{n}}$
That is, coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ is twice the coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n-1}}$.
Hence, proved.
Therefore, the correct answer is option (a).
Note: In this question, we just need a coefficient of one of the terms, which is, ${{x}^{n}}$. So, instead of writing whole binomial expansion, we can directly write coefficient on ${{x}^{n}}$ using the formula for ${{r}^{th}}$ term of binomial expansion of ${{\left( a+b \right)}^{n}}$, which is given by,
\[{{r}^{th}}\text{ term =}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\].