${(1 + x)}^{n} = p_{0} + p_{1} x + p_{2} x^{2} + \dots + p_{n} x^{n}, then p_{0} + p_{3} + p_{6} + \dots \dots =$

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$\frac{1}{3} [2^{n - 1} + \cos \frac{nπ}{3}]$

$\frac{2}{3} [2^{n - 1} + \cos \frac{nπ}{3}]$

$\frac{1}{3} [2^{n - 2} + \sin \frac{nπ}{3}]$

$\frac{2}{3} [2^{n - 2} + \sin \frac{nπ}{6}]$

answer is B.

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Detailed Solution

${(1 + x)}^{n} = p_{0} + p_{1} x + p_{2} x^{2} + . . . . + p_{n} x^{n}$

Sub $x = 1, w, w^{2}$

$\begin{array}{rcl} 2^{n} & = & p_{0} + p_{1} + p_{2} + . . . . + p_{n} \end{array}$

$\begin{array}{rcl} {(1 + w)}^{n} & = & p_{0} + p_{1} w + p_{2} w^{2} + . . . . \end{array}$

$\begin{array}{rcl} {(1 + w^{2})}^{n} & = & p_{0} + p_{1} w^{2} + p_{2} w + . . . . \end{array}$

$\begin{array}{rcl} ∴ 2^{n} + {(1 + w)}^{n} + {(1 + w^{2})}^{n} & = & 3 (p_{0} + p_{3} + p_{6} + . . . .) \end{array}$

$\begin{array}{rcl} 2^{n} + {(1 - \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{n} + {(1 - \frac{1}{2} - \frac{i \sqrt{3}}{2})}^{n} & = & 3 (p_{0} + p_{3} + p_{6} + . . . .) \end{array}$

$\begin{array}{rcl} p_{0} + p_{3} + p_{6} + . . . & = & \frac{1}{3} [2^{n} + {(\frac{1}{2} + \frac{i \sqrt{3}}{2})}^{n} + (\frac{1}{2} - \frac{i \sqrt{3}}{2})] \\ = & \frac{1}{3} (2^{n} + {(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{n} + {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{n}) \\ = & \frac{1}{3} (2^{n} + \cos n \frac{π}{3} + i \sin n \frac{π}{3} + \cos n \frac{π}{3} - i \sin n \frac{π}{3}) \\ = & \frac{1}{3} (2^{n} + 2 \cos n \frac{π}{3}) \end{array}$