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what is the value:(cos pi/2+isinpi/2)(cospi/2 square+isinpi/2 suqare)(cospi/2 cube+isinpi/2cube)...upto infinity

TARUN , 8 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 8 Years ago

Dear Tarun
 
When you put the de moivre theorem,
You get
cos(\pi/2 + \pi/2² + \pi/2³+.....) + i sin (\pi/2 + \pi/2² + \pi/2³+.....)
 
Since bracket is a sum of infinite terms of G.P.
Hence
Sum = (\pi/2)/(1- 1/2) = \pi
Now
The expression becomes as
Cos  \pi + i sin \pi
= -1 + 0
= -1
 
Regards
Arun (askIITians forum expert)
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