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Question Number 164071 by kdaramaths last updated on 13/Jan/22

Answered by Ar Brandon last updated on 13/Jan/22

Z=Σ_(k=0) ^n e^(ikx) =((1−e^(i(n+1)x) )/(1−e^(ix) )) =((e^(i((((n+1)x)/2))) (e^(−i((((n+1)x)/2))) −e^(i((((n+1)x)/2))) ))/(e^(i(x/2)) (e^(−i(x/2)) −e^(i(x/2)) ))) =e^(i((nx)/2)) ((sin((((n+1)x)/2)))/(sin((x/2))))=((sin((((n+1)x)/2)))/(sin(x/2)))[cos((nx)/2)+isin((nx)/2)] Σ_(k=0) ^n sinkx=ImΣ_(k=0) ^n e^(ix) =((sin((((n+1)x)/2))sin(((nx)/2)))/(sin((x/2)))) =((4sin((((n+1)x)/2))sin(((nx)/2))sin((x/2)))/(4sin^2 (x/2))) =((2sin((((n+1)x)/2))[cos((((n−1)x)/2))−cos((((n+1)x)/2))])/(2(1−cosx))) =(((sin(nx)+sinx)−(sin((n+1)x))/(2−2cosx)) =((sin((n+1)x)−sin(nx)−sinx)/(2cosx−2))

Z=∑nk=0eikx=1−ei(n+1)x1−eix=ei((n+1)x2)(e−i((n+1)x2)−ei((n+1)x2))eix2(e−ix2−eix2)=einx2sin((n+1)x2)sin(x2)=sin((n+1)x2)sinx2[cosnx2+isinnx2]∑nk=0sinkx=Im∑nk=0eix=sin((n+1)x2)sin(nx2)sin(x2)=4sin((n+1)x2)sin(nx2)sin(x2)4sin2x2=2sin((n+1)x2)[cos((n−1)x2)−cos((n+1)x2)]2(1−cosx)=(sin(nx)+sinx)−(sin((n+1)x)2−2cosx=sin((n+1)x)−sin(nx)−sinx2cosx−2

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