Question

cho:S=1+x+x²+x³+x⁴+x⁵

chứng minh;xS-S=x⁶-1

Answer 1

Ta có: \(x.S - S = x\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5}} \right) - \left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5}} \right)\)

\(\begin{array}{l} = x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6} - 1 - x - {x^2} - {x^3} - {x^4} - {x^5}\\ = {x^6} - 1 \text{(đpcm)} \end{array}\)

Answer 2

Ta có: \(S=1+x+x^2+x^3+x^4+x^5\)

\(x\cdot S=x\left(1+x+x^2+x^3+x^4+x^5\right)=x+x^2+x^3+x^4+x^5+x^6\)

Do đó: \(x\cdot S-S=\left(x+x^2+x^3+x^4+x^5+x^6\right)-\left(1+x+x^2+x^3+x^4+x^5\right)\)

\(=x+x^2+x^3+x^4+x^5+x^6-1-x-x^2-x^3-x^4-x^5\)

\(=x^6-1\)(đpcm)

Answer 3

\(S=1+x+x^2+x^3+x^4+x^5\\ \Rightarrow xS=x\left(1+x+x^2+x^3+x^4+x^5\right)\\ xS=x+x^2+x^3+x^4+x^5+x^6\\ \Rightarrow xS-S=\left(x+x^2+x^3+x^4+x^5+x^6\right)-\left(1+x+x^2+x^3+x^4+x^5\right)\\ xS-S=x^6-1\)