Question

A= 1+3+3²+3³+...+3⁹⁹

Answer 1

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

2A=\(3^{100}-1\)

A=\(\frac{3^{100}-1}{2}\)

Answer 2

 A=1+3+3^2+3^3+...+3^99

3A=3.(3^0+3^1+3^2+3^3+...+3^99)

3A=3^1+3^2+3^3+3^4+...+3^99+3^100

3A-A=2A=3^100-1

\          A=(3^100-1)/2

Chắc chắn đúng!!!