If Cr=Cr n then C0+C4+C8+C12+……=

(1+x)n=C0+C1x+C2x2+....CnxnPut x=1, -1, i, -i2n=C0+C1+C2+....Cn0=C0-C1+C2+....(C)n Cn(1+i)n=C0+C1 i-C2-C3 i+C4+....(1-i)n=C0-C1 i-C2+C3 i+C4+....add2n+(1+i)n+(1-i)n=4C0+C4+C8+.....(1+i)n+(1-i)n=2n cos nπ4+i sin nπ4+cos nπ4-i sin nπ4 =2n/2 2cos nπ4C0+C4+C8+.......=2n+2n2 2cos nπ44 =2n/2 2n/2-1+cos nπ42

If Cr=Cr n then C0+C4+C8+C12+……=

(1+x)n=C0+C1x+C2x2+....CnxnPut x=1, -1, i, -i2n=C0+C1+C2+....Cn0=C0-C1+C2+....(C)n Cn(1+i)n=C0+C1 i-C2-C3 i+C4+....(1-i)n=C0-C1 i-C2+C3 i+C4+....add2n+(1+i)n+(1-i)n=4C0+C4+C8+.....(1+i)n+(1-i)n=2n cos nπ4+i sin nπ4+cos nπ4-i sin nπ4 =2n/2 2cos nπ4C0+C4+C8+.......=2n+2n2 2cos nπ44 =2n/2 2n/2-1+cos nπ42

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If $C_{r} = {}^{n}C_{r}$ then $C_{0} + C_{4} + C_{8} + C_{12} + \dots \dots =$

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

$2^{\frac{n}{2}} \sin \frac{nπ}{4}$

$2^{n - 1} \cos \frac{nπ}{4}$

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

answer is D.

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Detailed Solution

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

Put $x = 1, - 1, i, - i$

$\begin{array}{rcl} 2^{n} & = & C_{0} + C_{1} + C_{2} + . . . . C_{n} \\ 0 & = & C_{0} - C_{1} + C_{2} + . . . . {(C)}^{n} C_{n} \\ {(1 + i)}^{n} & = & C_{0} + C_{1} i - C_{2} - C_{3} i + C_{4} + . . . . \\ {(1 - i)}^{n} & = & C_{0} - C_{1} i - C_{2} + C_{3} i + C_{4} + . . . . \end{array}$

add

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

$\begin{array}{rcl} {(1 + i)}^{n} + {(1 - i)}^{n} & = & {(\sqrt{2})}^{n} [\cos \frac{nπ}{4} + i \sin \frac{nπ}{4} + \cos \frac{nπ}{4} - i \sin \frac{nπ}{4}] \end{array}$

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