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Q. The coefficient of $x^{n}$ in the polynomial $\left(\right. x + \,{}^{2 n + 1} C_{0} \left.\right) \left(\right. x + \,{}^{2 n + 1} C_{1} \left.\right) \left(\right. x + \,{}^{2 n + 1} C_{2} \left.\right) ..... \left(\right. x + \,{}^{2 n + 1} C_{n} \left.\right)$ is

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

Given Expression is, $\left(\right. x + \,{}^{2 n + 1} C_{0} \left.\right) \left(\right. x + \,{}^{2 n + 1} C_{1} \left.\right) \left(\right. x + \,{}^{2 n + 1} C_{2} \left.\right) ..... \left(\right. x + \,{}^{2 n + 1} C_{n} \left.\right)$
If $P$ is coefficient of $x^{n}$ then,
$P = \,{}^{2 n + 1} C_{0} + \,{}^{2 n + 1} C_{1} + \,{}^{2 n + 1} C_{2} + ..... + \,{}^{2 n + 1} C_{n}$ -------(1)
$\Rightarrow P = \,{}^{2 n + 1} C_{2 n + 1} + \,{}^{2 n + 1} C_{2 n} + \,{}^{2 n + 1} C_{2 n - 1} + ..... + \,{}^{2 n + 1} C_{n + 1}$ ------(2) $\left(\right. \because \,{}^{n} C_{r} =^{n} C_{n - r} \left.\right)$
adding (1) and (2)
$2 P = \left(\right. ^{2 n + 1} C_{0} +^{2 n + 1} C_{1} + ....... +^{2 n + 1} C_{2 n + 1} \left.\right)$
$2 P = 2^{2 n + 1}$
$\therefore \, \, \, P=2^{2 n}$