Question

If $C_0,C_1,\cdots C_n$ are the coefficient of $x$ in expansion of $(1+x{)}^n$, then $C_0-C_2+C_4-C_6+\cdots +(-1{)}^n{C}_n=$

• A. $(2{)}^{n/2}\cos \frac{n\pi }{4}$
• B. $(2{)}^{n/2}\sin \frac{n\pi }{4}$
• C. $(2{)}^n\sin \frac{n\pi }{4}$
• D. $(2{)}^n\cos \frac{n\pi }{4}$

Answer 1

The correct option is A $(2{)}^{n/2}\cos \frac{n\pi }{4}$

Consider
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n$
Put $x=i$ where $i=\sqrt{-1}$
$(1+i{)}^n=C_0+C_{1}i+C_2{i}^2+C_3{i}^3+......$
$=(C_0-C_2+C_4-C_6)+i(C_1-C_3+C_5)$
$\Rightarrow C_0-C_2+C_4-C_6+....=\text{Real part of}(1+i{)}^n$
$=Re\left[\right(\sqrt{2}{)}^n{(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})}^n]$
$=Re\left[\right(\sqrt{2}{)}^n{(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})}^n]$
$=Re\left[\right(\sqrt{2}{)}^n(\cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4})]$(De' moivre's theorem)
$=(2{)}^{n/2}\cos \frac{n\pi }{4}$