Friday, November 30, 2018

calculus - Find the following finite sum


f(x)=c2014x2014+c2013x2013++c1x+c0 has 2014 roots a1,,a2014 and g(x)=c2014x2013+c2013x2012++c1. Given that c2014=2014 and f(x) is the derivative of f(x), find the sum 2014n=1g(an)f(an).





f(x)=2014(xa1)(xa2)(xa2014)



f(x)=20142x2013+2013c2013x2012++c1



is there any relation between f(an), g(an) and f(an) or do you need different approach to solve?
Edit
As Gerry suggested
now I have

c02014(1a1i1(a1ai)+1a2i2(a2ai)++1a2014i2014(a2014ai))



Would be helpful if someone tell me what to do next

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