$If C_{r} = {n_{C}}_{r}, then C_{0} + C_{4} + C_{8} + C_{12} = . . . . . =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2^{\frac{n}{2}} [\sin \frac{nπ}{4} + 2^{\frac{n}{2} - 1}]}{2}$

$2^{\frac{n}{2}} \sin \frac{nπ}{4}$

$2^{n - 1} \cos \frac{nπ}{4}$

$\frac{2^{\frac{n}{2}} [\cos \frac{nπ}{4} + 2^{\frac{n}{2} - 1}]}{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

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

$\begin{array}{rcl} c_{0} + c_{2} + c_{4} + . . . . . & = & 2^{n - 1} \\ c_{0} - c_{2} + c_{4} + . . . . . & = & 2^{\frac{n}{2}} \cos n \frac{π}{4} \\ 2 (c_{0} + c_{4} + . . . . .) & = & 2^{n - 1} + 2^{\frac{n}{2}} \cos n \frac{π}{4} \\ c_{0} + c_{4} + c_{8} + . . . . . & = & \frac{2^{\frac{n}{2}} [2^{\frac{n}{2} - 1} + \cos n \frac{π}{4}]}{2} \end{array}$