Question

somebody helps me please ( mik cần gấp lắm)

Cho A= 3+3²+3³+3⁴+...+3¹⁰⁰. Tìm x thuộc N để 2.A+3=3^x

Ai nhanh nhất mik tick nhé

Answer 1

\(A=3+3^2+3^3+3^4+...+3^{100}\)

\(3A=3^2+3^3+...+3^{101}\)

\(3A-A=\left[3^2+3^3+...+3^{101}\right]-\left[3+3^2+3^3+...+3^{100}\right]\)

\(2A=3^{101}-3\)

\(A=\frac{3^{101}-3}{2}\)

Ta lại có : \(2A+3=3^x\)

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

=> \(3^{101}-3+3=3^x\)

=> 3¹⁰¹ = 3^x

=> x = 101

Vậy x = 101

Answer 2

\(3A=3^2+3^3+...+3^{101}\)

\(\Rightarrow2A=3^{101}-3\)

\(\Rightarrow2A+3=3^{101}=3^x\)

\(\Rightarrow x=101\)