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]
Subtracting equations [3] from [2]:
(1+i)n−(1−i)n=2i[P1−P3+P5.........]
⟹[√2ei(π/4)]n−[√2e−i(π/4)]n=2i[P1−P3+P5.........]
⟹(2)n2ei(nπ/4)−(2)n2e−i(nπ/4)=2i[P1−P3+P5.........]
⟹(2)n2[cos n(π/4)+isin n(π/4)−cos n(π/4)+sin n(π/4)]=2i[P1−P3+P5.........]
⟹(2)n2[2isin nπ4]=2i[P1−P3+P5........]
Hence,
⟹(2)n2[sin nπ4]=[P1−P3+P5........]
So, m=4.