Question

Tìm x biết:

\(B=3+3^2+3^3+...+3^{2006}và2B+3=3^x\)

Answer 1

B = 3 + 3² + 3³ + ... + 3²⁰⁰⁶

2B + 3 = 3B - B + 3 = (3² + 3³ + 3⁴ + ... + 3²⁰⁰⁷) - (3 + 3² + 3³ + ... + 3²⁰⁰⁶) + 3 = 3²⁰⁰⁷ - 3 + 3 = 3²⁰⁰⁷ = 3^x => x = 2007

Answer 2

\(B=3+3^2+3^3+.....+3^{2006}\)

\(\Rightarrow3B=3^2+3^3+3^4+.....+3^{2007}\)

\(\Rightarrow2B=3^{2007}-3\)

\(\Rightarrow B=\frac{3^{2007}-3}{2}\)

\(2B+3=3^x\)

\(\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\)

\(\Rightarrow3^{2007}-3+3=3^x\)

\(\Rightarrow3^{2007}=3^x\)

\(\Rightarrow x=2007\)

Answer 3

B=3+3^2+3^3+...+3^2006 =>3B=3^2+3^3+...+3^2006+3^2007

=>2B=3^2007-3

=>2B+3=3^2007-3+3=3^2007=>x=2007

nhớ k cho mk nha