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Question : 18 of 160
Marks:
+1,
-0
Solution:
It is given that for a complex number
z,
(1+z)n=1+nC1z+nC2z2+nC3z3+...+nCnzn
Put
z=cosx+isinx Then,
(1+cosx+isinx)n=1+nC1(cosx+isinx)
+nC2(cosx+isinx)2+...+nCn(cosx+isinx)n
⇒2n(cos)n(cos+isin)n =[1+nC1cosx+nC2cos(2x)+nC3cos(3x)
+...+nCncos(nx)]+i[nC1sinx+nC2sin(2x)
+nC3sin(3x)+...+nCnsin(nx)] ⇒2n(cos)n(cos+isin) =[1+nC1cosx+nC2cos(2x)+nC3cos(3x)
+...+nCncos(nx)]+i[nC1sinx+nC2sin(2x)
+nC3sin(3x)+...+nCnsin(nx)] On comparing imaginary parts, we get
nC1sinx+nC2sin(2x)+nC3sin(23x) +...+nCnsin(nx =2n(cos)nsin() =(2cos)100sin(kx)( given
) Therefore,
k=50.
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