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Question

Series 1+cosα+1+cos2α+1+cos3α+.......... to n terms.

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Solution

1+cosα+1+cos3α+.....+1+cos3α+....+1+cosnα
Using 2cos2x2=1+cosx
2[cosα2+cos2α2+cos3α2+.....+cosnα2]........(1)
Using n1k=0cos(a+kd)=sin(n d/2)sin(d/2)cos(2a+(n1)d2)
Here, a=α2,d=d2
(1)
2⎢ ⎢ ⎢ ⎢sin(dα4)sin(α4)cos(α4(n+1))⎥ ⎥ ⎥ ⎥

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