Question

If $(1+x{)}^n=P_0+P_{1}x+P_2{x}^2+P_n{x}^n$ and $\theta =\frac{n\pi }{4}$, then what will be the value of the expression $P_0-P_3+P_5+....-\left(2^{\frac{n}{2}}sin\theta \right)$ ?

Answer 1

Given,

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

Putting $x=i$ in eq.[1] :-

$(1+i{)}^n=P_0+P_1(i)+P_2(i{)}^2..........+P_n(i{)}^n$

$⟹(1+i{)}^n=P_0+i{P}_1-P_2-i{P}_3+P_4..........P_n(i{)}^n$..............................[2]

Now putting $x=-i$ in eq.[1] :-

$(1-i{)}^n=P_0+P_1(-i)-P_2+i{P}_3+P_4...............P_n(-i{)}^n$

$⟹(1-i{)}^n=P_0-P_1+P_2.............+(-1{)}^n{P}_n{x}^n$.............................[3]

Subtracting equations $\left[3\right]$ from $\left[2\right]$:

$(1+i{)}^n-(1-i{)}^n=2i[P_1-P_3+P_5.........]$

$⟹[\sqrt{2}{e}^{i\left(\pi /4\right)}{]}^n-[\sqrt{2}{e}^{-i\left(\pi /4\right)}{]}^n=2i[P_1-P_3+P_5.........]$

$⟹(2{)}^{\frac{n}{2}}{e}^{i\left(n\pi /4\right)}-(2{)}^{\frac{n}{2}}{e}^{-i\left(n\pi /4\right)}=2i[P_1-P_3+P_5.........]$

$⟹\left(2{)}^{\frac{n}{2}}\right[cosn\left(\pi /4\right)+isinn\left(\pi /4\right)-cosn\left(\pi /4\right)+sinn\left(\pi /4\right)]=2i[P_1-P_3+P_5.........]$

$⟹\left(2{)}^{\frac{n}{2}}\right[2isin\frac{n\pi }{4}]=2i[P_1-P_3+P_5........]$

Hence,

$⟹\left(2{)}^{\frac{n}{2}}\right[sin\frac{n\pi }{4}]=[P_1-P_3+P_5........]$

So, $[P_1-P_3+P_5........(2{)}^{\frac{n}{2}}\left[sin\frac{n\pi }{4}\right]]=0$