If ( 1 + x ) n = p 0 + p 1 x + p 2 x 2 + … + p n x n , then p 0 + p 3 + p 6 + … =

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If $(1+x)^n=p_0+p_1 x+p_2 x^2+\ldots+p_n x^n$, then $p_0+p_3+p_6+\ldots=$

MathematicsComplex NumberAP EAMCETAP EAMCET 2017 (24 Apr Shift 2)

Options:

A $\frac{1}{3}\left[2^{n-1}+\cos \frac{n \pi}{3}\right]$
B $\frac{2}{3}\left[2^{n-1}+\cos \frac{n \pi}{3}\right]$
C $\frac{1}{3}\left[2^{n-2}+\sin \frac{n \pi}{3}\right]$
D $\frac{2}{3}\left[2^{n-2}+\sin \frac{n \pi}{6}\right]$

Solution:

1945 Upvotes Verified Answer

The correct answer is: $\frac{2}{3}\left[2^{n-1}+\cos \frac{n \pi}{3}\right]$

No solution. Refer to answer key.

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