Question

 Tìm x biết :x=1+3+\(3^2+3^3\)+...+3\(^{100}\)

Answer 1

x=1+3+3²+...+3¹⁰⁰

=>3x=3+3²+3³+...+3¹⁰¹

=>3x-x=(3+3²+3³+...+3¹⁰¹)-(1+3+3²+...+3¹⁰⁰)

=>2x=3¹⁰¹-1

=>x=\(\frac{3^{101}-1}{2}\)

Vậy x=\(\frac{3^{101}-1}{2}\)

Answer 2

\(x=1+3+3^2+...+3^{100}\Rightarrow3.x=3+3^2+3^3+...+3^{100}+3^{101}\)

\(\Rightarrow3.x-x=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)=3^{101}-1\)

\(\Rightarrow2.x=3^{101}-1\Rightarrow x=\frac{3^{101}-1}{2}\)