已知x+1x=2，求代数式x+x2+x4+x8+…+x1024+1x+1x2+1x4+1x8+…+1x1024．

Question

已知x+(1)/(x)=2，求代数式x+x²+x⁴+x⁸+…+x¹⁰²⁴+(1)/(x)+(1)/(x2)+(1)/(x4)+(1)/(x8)+…+(1)/(x1024)．

Answer 1

∵x+(1)/(x)=2，
∴x2+(1)/(x2)=(x+(1)/(x))2-2=2²-2=2，
∴x4+(1)/(x4)=(x2+(1)/(x2))2-2=2²-2=2，
∴x+x²+x⁴+x⁸+…+x¹⁰²⁴+(1)/(x)+(1)/(x2)+(1)/(x4)+(1)/(x8)+…+(1)/(x1024)
=(x+(1)/(x))+(x2+(1)/(x2))+…+(x1024+(1)/(x1024))
=

2+2+2+…+2

11个2

=2×11
=22．