Let n is of the form of 3P where P is an odd integer then

$^{n} C_{0}$ + $^{n} C_{3}$ + $^{n} C_{6}$ + $^{n} C_{9}$ + ........ + $^{n} C_{n}$ equals

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Solution

The correct option is A

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

$(1 + ω)^{n} = c_{0} + c_{1} ω + c_{2} ω^{2} + . . . . . . . . + c_{n} ω^{n}$

$(1 + ω^{2})^{n} = c_{0} + c_{1} ω^{2} + c_{2} ω^{4} + . . . . . . . . + c_{n} ω^{2 n}$

$2^{n} = c_{0} + c_{1} + c_{2} + . . . . . + c_{n}$

$2^{n} + (- ω)^{n} + (- ω^{2})^{n} = 3 c_{0} + 3 c_{3} + . . . . + 3^{n} C_{n}$

$c_{0} + c_{3} + c_{6} + . . . . . . + c_{n} = \frac{1}{3} [26 n + (- 1)^{n} ω^{n} + (- 1)^{n} ω^{2 n}]$

= $\frac{1}{3} [2^{n} + (- 1)^{3 P} ω^{3 P} + (- 1)^{3 P} ω^{6 P}]$

= $\frac{1}{3} [2^{n} - 1 - 1] = \frac{1}{3} [2^{n} - 2]$

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