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Question

The value of sinπn+sin3πn+sin5πn+ to n terms is

A
0
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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 A 0
We know that,
sinA+sin(A+D)++sin(A+(n1)D)=sinnD2sinD2[sin(2A+(n1)D2)]
Now, using the above fornula
sinπn+sin3πn+sin5πn+ to n terms=sinn2π2nsin2π2n⎢ ⎢ ⎢sin⎜ ⎜ ⎜2πn+(n1)2πn2⎟ ⎟ ⎟⎥ ⎥ ⎥=0

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