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(x−a1)(x−a2)…(x−a2014)
f′(x)=20142x2013+2013⋅c2013x2012+⋯+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(1a1∏i≠1(a1−ai)+1a2∏i≠2(a2−ai)+⋯+1a2014∏i≠2014(a2014−ai))
Would be helpful if someone tell me what to do next
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