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Prove that sum(r=0)^(n) ""^(n)C(r).(n-r)...

Prove that `sum_(r=0)^(n) ""^(n)C_(r).(n-r)cos((2rpi)/(n)) = - n.2^(n-1).cos^(n)'(pi)/(n)`.

Text Solution

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`S = underset(r=0)overset(n)sum.^(n)C_(r) . (n-r) cos'((2rpi)/(n))"……"(1)`
` = underset(r=0)overset(n)sum.^(n)C_(n-r).(n-(n-r))cos((2(n-r)pi)/(n))`
` :. S = underset(r=0)overset(n)sum.^(n)C_(r).cos((2rpi)/(n)) "……."(2)`
Adding (1) and (2), we get
`2S = n underset(r=0)overset(n)sum.^(n)C_(r).cos((2rpi)/(n)) = n xx Re(underset(r=0)overset(n)sum.^(n )C_(r)e^(i(2rpi)/(n)))`
`= n xx Re (1+e^((2pi)/(n)i))^(n)`
` = n xx Re (1+cos'(2pi)/(n)+isin'(2pi)/(n))^(n)`
`= nxx Re(2cos^(2)'(pi)/(n)+2isin'(pi)/(n) cos'(pi)/(n))^(n)`
`= n2^(n)cos^(n)'(pi)/(n) Re(cos'(pi)/(n) + isin'(pi)/(n))^(n)`
`= n2^(n)cos^(n)'(pi)/(n)Re(cos'(npi)/(n)+isin'(npi)/(n))`
`:. S = - n2^(n-1)cos^(n)'(pi)/(n)`
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