1The answer is wrong.The answer I have is n.2n-1cosn-1@2.cos(n-1)@2.
The value of C1+2C2cos@+3C3cos2@+4C4cos3@+.................................+nCncos[(n-1)@]
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4 Answers
1( 1 + x ) n = 1 + n C 1 x + n C 2 x 2 + ................+ n C n x n
Differentiating with respect to " x " ,
n ( 1 + x ) n - 1 = n C 1 + 2 n C 2 x + ................ + n n C n x n - 1 ....................... ( 1 )
Now ,
1 + e i θ = 1 + cos θ + i sin θ = 2 cos 2 θ + 2 i sin θ2 cos θ2
= 2 cos θ2 ( cos θ2 + i sin θ2 ) = 2 cos θ2 e i θ2
From ( 1 ) , putting " x = e i θ " and equating real parts ,
n C 1 + 2 n C 2 cos θ + ................ + n n C n cos ( n - 1 ) θ = 2 n cos n θ2 cos n θ2
11Yup , if you put " e i θ " in ( 1 ) , then you get the result you stated . I actually put " e i θ " in " ( 1 + x ) n " .