Question

The term independent of x in the expansion of$(1 + x)^{n}\left( 1 + \frac{1}{x} \right)^{n}$ is

• A. $C_{0}^{2} + 2C_{1}^{2} + ...... + (n + 1)C_{n}^{2}$
• B. $(C_{0} + C_{1} + ...... + C_{n})^{2}$
• C. $C_{0}^{2} + C_{1}^{2} + ...... + C_{n}^{2}$
• D. None of these

Answer 1

Answer: (C)

$C_{0}^{2} + C_{1}^{2} + ...... + C_{n}^{2}$

Answer 2

We know that,

$\left( 1 + \frac{1}{x} \right)^{n} =^{n} ⥂ C_{0} +^{n} ⥂ C_{1}\frac{1}{x^{1}} +^{n} ⥂ C_{2}\frac{1}{x^{2}} + ..... +^{n} ⥂ C_{n}\frac{1}{x^{n}}$Obviously, the

term independent of x will be

$n ⥂ C_{0}.^{n} ⥂ C_{0} +^{n} ⥂ {C_{1}}^{n} ⥂ C_{1} + ..... +^{n} ⥂ C_{n}.^{n} ⥂ C_{n} = C_{0}^{2} + C_{1}^{2} + ....... + C_{n}^{2}$

Trick : Put $n = 1$ in the expansion of

$(1 + x)^{1}\left( 1 + \frac{1}{x} \right)^{1} = 1 + x + \frac{1}{x} + 1 = 2 + x + \frac{1}{x}$.....(i)

We want coefficient of $x^{0}$. Comparing to equation (i).

Then, we get 2 i.e., independent of x.

Option (3) : $C_{0}^{2} + C_{1}^{2} + .....C_{n}^{2}$; Put $n = 1$;

Then $1C_{0}^{2} +^{1} ⥂ C_{1}^{2} = 1 + 1 = 2$.