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Questions  

If f(x)=5x5+5x, then the value of f120+f220++f3920 is

a
20
b
292
c
192
d
392

detailed solution

Correct option is D

We have,f(x)=5x5+5x⇒f(2−x)=52−x5+52−x=255x+1+52=55x+5∴ f(x)+f(2−x)=1 for all xNow,f120+f220+…..f3820+f3920=f120+f3920+f220+f3820+…+f1920+f2120+f2020=1×19+f(1)=19+55+5=392

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