Let f be a polynomial function such that log2(f(x))=(log2(2+23+29+...∞))·log3(1+f(x)f(1/x)), x > 0 and f(6) = 37. Then ∑n=110f(n) is equal to ________. [2026]
(395)
S=2+23+29+...∞
S=a1–r ⇒ S=21–1/3=3
log2(f(x)=log2(3)·log3(1+f(x)f(1/x))
Using base change formula
log2(3)·log3(A)=log2(A)
⇒ log2(f(x)=log2(1+f(x)f(1/x))
⇒ f(x)=1+f(x)f(1/x)
⇒ f(x)–1=f(x)f(1/x)
⇒ f(1/x)=f(x)f(x)–13
⇒ f(x)·f(1/x)=f(x)+f(1/x)
⇒ f(x)=1±xn ⇒ f(6)=37
If f(x)=1–xn, then 1–6n=37 ⇒ –6n=36 (no real solution)
If f(x)=1+xn, then 1+6n=37 ⇒ 6n=36 ⇒ n=2
f(x)=x2+1
∑n=110f(n)=∑n=110(n2+1)=∑n=110n2+∑n=1101
=n(n+1)(2n+1)6+10⏟for n=10
=10×11×216+10=385+10=395