Answer: Option B
:
B
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.........+C_n{X}^n$ ...........(i)
and $(1+\frac{1}{x}{)}^n$ =$C_0+C_1\frac{1}{x}+C_2(\frac{1}{x}{)}^2+........+C_n(\frac{1}{x}{)}^n$ ..........(ii)
If we multiply (i) and (ii), we get
$C_0^2+C_1^2+C_2^2+........+C_n^2$
is the term independent of x and hence it is equal to the term independent of x in the product $(1+x{)}^n(1+\frac{1}{x}{)}^n$ or in $\frac{1}{x^n}$$(1+x{)}^{2n}$ or term
containing $x^n$ in $(1+x{)}^{2n}$. Clearly the coefficient of $x^n$ in $(1+x{)}^{2x}$ is $T_{n+1}$ and equal to${}^{2n}{C}_n$ = $\frac{\left(2n\right)!}{n!n!}$
Trick:Solving conversely.
Put n = 1, n = 2,.........then we get $S_1$ = ${}^1{C}_0^2{+}^1{C}_1^2$= 2,
$S_2$ =${}^2{C}_0^2{+}^2{C}_1^2{+}^2{C}_2^2=1^2+2^2+1^2$ = 6
Now check the options
(a) Does not hold given condition,
(b) (i) Put n = 1, then $\frac{2!}{1!1!}=2$
(ii) Put n = 2, then $\frac{4!}{2!2!}=\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=6$
Note: Students should remember this question as identity