Question

A. ${}^{2n}{C}_n=C_0^2+C_1^2+C_2^2+C_3^2+\dots \cdots +C_n^2$
B. ${}^{2n}{C}_n=$ term independent of $x$ in $(1+x{)}^n{(1+\frac{1}{x})}^n$
C. ${}^{2n}{C}_n=\frac{\mathrm{1.3.5.7}\dots \dots (2n-1)}{n!}$ then

• A. A, B are false, C is true
• B. A is false, B and C are true
• C. A and B are true; C is false
• D. A, B, C are true

Answer 1

The correct option is C A and B are true; C is false

${}^{2n}{C}_n$ $=\frac{2n!}{(n!{)}^2}$

$=\frac{2^n\left(\mathrm{1.3.5..}\right(2n-1\left)\right)}{n!}$

Hence option C is false.

Consider Option B
$(1+x{)}^n\cdot (1+\frac{1}{x}{)}^n$
$=\frac{(1+x{)}^n(1+x{)}^n}{x^n}$
$=\frac{(1+x{)}^{2n}}{x^n}$.

The general term in this case is
$T_{r+1}={}^{2n}{C}_r{x}^r.x^{-n}$
$={}^{2n}{C}_r{x}^{r-n}$
For term independent of $x$
$r-n=0$
$n=r$.

Hence the term independent of $x$ will be
${}^{2n}{C}_n$
Thus B is true.

Consider option A.
Consider the following series.
${}^m{C}_0{}^n{C}_r+{}^m{C}_1{}^n{C}_{r-1}+...{}^m{C}_r{}^n{C}_0$
$=$ Coefficient of $x^r$ in $(1+x{)}^m.(x+1{)}^n$
$=$ Coefficient of $x^r$ in $(1+x{)}^{m+n}$
$={}^{m+n}{C}_r$

In the above case, $r=n$ and $m=n$
Hence, ${}^n{C}_0^2+{}^n{C}_1^2+{}^n{C}_2^2+{}^n{C}_3^2+..{}^n{C}_n^2$
$={}^n{C}_0{}^n{C}_n+{}^n{C}_1{}^n{C}_{n-1}+...{}^n{C}_2{}^n{C}_{n-2}$
$={}^{2n}{C}_n$

Hence A is also true.