If f(x)=5x5+5x, then the value of f120+f220+…+f3920 is
20
292
192
392
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 x
Now,
f120+f220+…..f3820+f3920=f120+f3920+f220+f3820+…+f1920+f2120+f2020=1×19+f(1)=19+55+5=392