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