Question

If ${(1+x)}^n=P_0+P_{1}x+P_2{x}^2+...+P_n{x}^n$,

then $\frac{p_0-p_2+p_4-p_6+...}{p_1-p_3+p_5-p_7+...}=?$

• A. $itan\frac{n\pi }{4}$
• B. $tan\frac{n\pi }{4}$
• C. $cot\frac{n\pi }{4}$
• D. $icot\frac{n\pi }{4}$

Answer 1

The correct option is C $cot\frac{n\pi }{4}$

Let $x=i$
Hence, $(1+i{)}^n=P_0+P_{1}i+P_2{i}^2+P_3{i}^3...+P_n{i}^n$
$(1+i{)}^n=P_0+P_{1}i-P_2-P_{3}i+...$
$(1+i{)}^n=(P_0-P_2+P_4-P_6+...)+i(P_1-P_3+P_5-...)$
$\Rightarrow (\sqrt{2}{)}^n.e^{\frac{in\pi }{4}}=(P_0-P_2+P_4+...)+i(P_1-P_3+P_5-...)$
Comparing real and imaginary parts,
$P_0-P_2+P_1...=(\sqrt{2}{)}^2\cos \frac{n\pi }{4}$------(1)
and $P_1-P_3+P_5....=(\sqrt{2}{)}^2\sin \frac{n\pi }{4}$----------(2)
So, $\frac{\left(1\right)}{\left(2\right)}=$ required expression $=cot\left(\frac{n\pi }{4}\right)$