Question

1. Tính H(x)= 1 +x+x²+x³ +.....+ x²⁰ tại x= 0,54321

Answer 1

\(H\left(x\right)=1+x+x^2+x^3+...+x^{20}\)

\(\Rightarrow xH\left(x\right)=x\left(1+x+x^2+x^3+...+x^{20}\right)\)

\(xH\left(x\right)x+x^2+x^3+x^4+...+x^{21}\)

\(\Rightarrow xH\left(x\right)-H\left(x\right)=\left(x-1\right)H\left(x\right)\left(x+x^2+x^3+x^4+...+x^{21}\right)-\left(1+x+x^2+x^3+...+x^{20}\right)\)

\(\left(x-1\right)H\left(x\right)=x^{21}-1\Leftrightarrow H\left(x\right)=\dfrac{x^{21}-1}{x-1}\)

thay \(x=0,54321\) vào \(H\left(x\right)\)

ta có : \(H\left(x\right)=\dfrac{0,54321^{21}-1}{0,54321-1}=2,18918383\) vậy \(H\left(x\right)=2,18918383\)