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Question

If Sn=sin2πn+sin22πn++sin2(n1)πn, then the value of S100 is 


Solution

Sn=sin2πn+sin22πn++sin2(n1)πn=12[1cos2πn+1cos4πn++1cos2(n1)πn]=12[(n1)(cos2πn+cos4πn++cos2(n1)πn)]

We know that,
(cos2πn+cos4πn++cos2(n1)πn)=cos0+cos2πn+cos(4πn)    ++cos(2(n1)πn)cos0=sinπsinπn⎢ ⎢ ⎢cos⎜ ⎜ ⎜2×0+(n1)2πn2⎟ ⎟ ⎟⎥ ⎥ ⎥1=1

Therefore,
Sn=12[n1(1)]=n2
S100=50

Mathematics

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