Find the value

find the imaginary part of

\large \sum_{n=1}^{90}{n \sin(2kn)}=imaginary(\sum_{n=1}^{90}{n e^{(2kn)i}})

the term on the right is a AG series. with first term common difference 2 and common ratio e^{2k i}

find the sum and then take its imaginary part.

there is no mistake..

your approach will give

180(sin 2+ sin 4+ sin 6.... sin 90)

you can either use the series sum for sin (a) + sin(a+d) ... sin (a+(n-1)d) from memory

or use the trick that i used above.

In fact your approach makes it even easier to find the sum i gave above.

The answer is double of what you are getting.

I think you have missed a 2 somewhere!

somehow 90 tan 89 is very close to the answer.. but not exactly equal.. !!!

this is by ur method...