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Question

If S = sinπn + sin3πn + sin5πn+..........n terms.

Find the value nS


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Solution

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 = sinnβ2sinβ2 sin(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

n.S = n × 0 = 0


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