The correct option is
A True
Given,
(1+x)n=P0+P1x+P2x2+Pnxn.............[1]
Putting x=i in eq.[1] :-
(1+i)n=P0+P1(i)+P2(i)2..........+Pn(i)n
⟹(1+i)n=P0+iP1−P2−iP3+P4..........Pn(i)n..............................[2]
Now putting x=−i in eq.[1] :-
(1−i)n=P0+P1(−i)−P2+iP3+P4...............Pn(−i)n
⟹(1−i)n=P0−P1+P2.............+(−1)nPnxn.............................[3]
Adding equations [2] and [3]:
(1+i)n+(1−i)n=2[P0−P2+P4.........]
⟹[√2ei(π/4)]n+[√2e−i(π/4)]n=2[P0−P2+P4.........]
⟹(2)n2ei(nπ/4)+(2)n2e−i(nπ/4)=2[P0−P2+P4.........]
⟹(2)n2[cos n(π/4)+isin n(π/4)+cos n(π/4)−sin n(π/4)]=2[P0−P2+P4.........]
⟹(2)n2[2cos nπ4]=2[P0−P2+P4........]
Hence,
⟹(2)n2[cos nπ4]=[P0−P2+P4........]
So, above statement is true.