Question

C=3+3^2+3^5+...+3^99

 

Answer 1

C=3+3²+3³+...+3⁹⁹

=>3C=3²+3³+...+3²⁰⁰⁶+3¹⁰⁰

=>3C-C=(3²+3³+...+3²⁰⁰⁶+3¹⁰⁰)-(3+3²+3³+...+3⁹⁹)

=>2C=3¹⁰⁰-3

=>C=\(\frac{3\left(3^{99}-1\right)}{2}\)

Answer 2

Ta có: C= 3 + 3² + 3³ + ... + 3⁹⁹

=>3C = 3² + 3³ +...+3⁹⁹ + 3¹⁰⁰

=>3C - C = 3¹⁰⁰ - 3

=>2C = 3¹⁰⁰-3

\(\Rightarrow C=\frac{3^{100}-3}{2}\)