How should i solve : n∑r=1(2r−1)cos(2r−1)θ
I can solve ∑nr=1cos(2r−1)θ by considering
ℜ∑nr=1z2r−1 and using summation of geometric series, but I can't seem to find a common geometric ratio when 2r−1 is involved in the summation.
Visually : ∑nr=1z2r−1=z+z3+...+z2r−1 where the common ratio r=z2 can easily be seen, but in the case of ∑nr=1(2r−1)z2r−1=z+3z3+5z5+...+(2r−1)z2r−1, how should i solve this ? A hint would be appreciated.
Answer
Note that n∑r=1(2r−1)cos(2r−1)θ=n∑r=1ddθsin(2r−1)θ=ddθn∑r=1sin(2r−1)θ
Now calculate n∑r=1sin(2r−1)θ=sin2(nθ)sin(θ) Through the formula for the sum of sin's.
Now just diffferentiate with regard to θ.
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