Lời giải:
Sử dụng công thức $(a-1)(a+1)=a^2-1$ ta có:
$8F=(9-1)(9+1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^2-1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^4-1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^8-1)(9^8+1)...(9^{32}+1)$
$=(9^{16}-1)...(9^{32}+1)=(9^{32}-1)(9^{32}+1)=9^{64}-1$
$\Rightarrow F=\frac{9^{64}-1}{8}$