Question

If $c_0,c_1,c_2\cdots c_n$ are binomial coefficients in ${(1+x)}^n$, then the value of $c_1+c_5+c_9+c_{13}+\cdots$ equals

• A. $2^{n-1}+2^{\frac{n}{2}}\sin \left(\frac{n\pi }{4}\right)$
• B. $2^{n-1}+2^{\frac{n}{2}}\cos \left(\frac{n\pi }{4}\right)$
• C. $\frac{1}{2}(2^{n-1}+2^{\frac{n}{2}}\sin \frac{n\pi }{4})$
• D. $\frac{1}{2}(2^{n-1}-2^{\frac{n}{2}}\sin \frac{n\pi }{4})$

Answer 1

The correct option is C $\frac{1}{2}(2^{n-1}+2^{\frac{n}{2}}\sin \frac{n\pi }{4})$

$c_0,c_1,c_2\cdots c_n$ are binomial coefficients of ${(1+x)}^n$
i.e ${(1+x)}^n=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\dots$
put $x=i$
$\Rightarrow {(1+i)}^n=c_0+c_{1}i-c_2-c_{3}i+\dots$
$\Rightarrow 2^{\frac{n}{2}}c$ is $\frac{n\pi }{4}=c_0+c_{1}i-c_2-c_{3}i+\dots$
Comparing imaginary parts gives
$2^{\frac{n}{2}}\sin \frac{n\pi }{4}=c_1-c_3+c_5+\dots$ ------(1)
$2^{n-1}=c_1+c_3+c_5+\dots$ ------(2) ($\because c_0+c_1+c_2+\cdots =2^n$)
$\therefore$ (1)+(2) gives
$c_1+c_5+c_9+\cdots =\frac{1}{2}(2^{n-1}+2^{\frac{n}{2}}\sin \frac{n\pi }{4})$