Question

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

giup mik với

 

Answer 1

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

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

3B = 3 + 3^2  ... 3^101

3B - B = 2B = (3 + 3^2  ... 3^101) - (1+3+3^2+3^3+..+3^100)

B = (3^101 - 1) : 2

Answer 2

Đặt A=1+3+3^2+3^3+...+3^100

\(\Rightarrow\)3A=3+3^2+3^3+...+3^101

Ta có : 3A-A=(3+3^2+3^3+...+3^101)-(1+3+3^2+3^3+..+3^100)

2A=3+3^2+3^3+...+3^101-1-3-3^2-3^3-..-3^100

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

Answer 3

3A=3¹+3²+3³+.........+3¹⁰¹     (1)

  A=3⁰+3¹+3²+.........+3¹⁰⁰     (2)

Ta lấy (1)-(2) => 2A=3¹⁰¹-3⁰=3¹⁰¹-1

Answer 4

À cho mình bổ sung thêm :

2A=3¹⁰¹-1 

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