Question

[vũ bảo ngọc]

TÍNH A=\(3^1+3^2+3^3+...+3^{99}+3^{100}\)

Answer 1

[Lê Tài Bảo Châu]

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

\(\Rightarrow3A=3^2+3^3+...+3^{101}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)

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

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

Answer 2

[Lê Nguyễn Quang Thắng]

Ta có: A = 3¹ + 3² + 3³ + … + 3⁹⁹+3¹⁰⁰

• 3A = 3.( 3 + 3² + 3³ + … + 3⁹⁹+3¹⁰⁰)
• 3A =  3² + 3³ + … + 3¹⁰⁰+3¹⁰¹
• 3A – A = (3² + 3³ + … + 3⁹⁹+3¹⁰⁰)-(3¹ + 3² + 3³ + … + 3⁹⁹+3¹⁰⁰)
• A = 3¹⁰⁰-3