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Question Number 54641 by Joel578 last updated on 08/Feb/19
Givenf(x)=4x+4x2−12x+1−2x−1Findthevalueoff(13)+f(14)+f(15)+...+f(112)
Commented by Meritguide1234 last updated on 08/Feb/19
f(x)=(2x−1)+(2x+1)+(2x+1)(2x−1)2x+1+2x−1⇒f(x)=denominatorshouldbepositive⇒f(x)=(2x−1)3−(2x+1)32f(13)+f(14)+...+f(112)=1625
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