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Question

If S=sinπn+sin3πn+sin5πn+...n terms, then nS= 


  1. \N
  2. 1
  3. 2
  4. 3


Solution

The correct option is A \N

Given series S=sinπn+sin3πn+sin5πn+...sin(α+(n1)β)

Comparing the given series with standard sine series:

sinα+sin(α+β)+sin(α+2β)+sin(α+3β)+...

α=πn

β=2πn

Sum of the sine series =sinnβ2sinβ2sin(α+(n1)β2)

=sin(n×2π2×n)sin(2π2×n)sin(πn+(n1)2π2.n)

=sinπsin(πn)×sin(πn+(n1)πn)

=0sin(πn)sin(πn+(n1)πn)

=0

nS=n×0=0

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