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B
1
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C
√3
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D
2
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Solution
The correct option is A0 We know that, sinA+sin(A+D)+⋯+sin(A+(n−1)D)=sinnD2sinD2[sin(2A+(n−1)D2)]
Now, using the above fornula sinπn+sin3πn+sin5πn+⋯ to n terms=sinn⋅2π2⋅nsin2π2⋅n⎡⎢
⎢
⎢⎣sin⎛⎜
⎜
⎜⎝2⋅πn+(n−1)⋅2πn2⎞⎟
⎟
⎟⎠⎤⎥
⎥
⎥⎦=0