Question

Tính: A= 1+3+3^2 +... +3^100 

Mong các bạn giúp mình với nha 

 

Answer 1

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

3A = 3 + 3 ^2 + 3^3 + ... +3^101

3A - A =  ( 3 + 3 ^2 + 3^3 + ... +3^101)

            -  ( 1 + 3 + 3^2 + ...+ 3 ^100)

2A = 3 ^101 - 1

A = \(\frac{\text{ 3 ^101 - 1}}{2}\)

Answer 2

Ta có: 

A= \(1+3+3^2+...+3^{100}\)

3A=\(3\times\left(1+3+3^2+...+3^{100}\right)\)

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

\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\)(1+3+3^2+...+3^100)

2A=1+3^101

A=(1+3^101)/2