Question

Tính tổng

S= 1+ 3 + 3² +... + 3¹⁸ + 3¹⁹

Giải giúp mk vs nha!!! 

Answer 1

S= 1+ 3 + 3² +... + 3¹⁸ + 3¹⁹

3S= 3+3²+3³+... + 3¹⁹+ 3²⁰

3S-S= ( 3+3²+3³+... + 3¹⁹+ 3²⁰)-(1+ 3 + 3² +... + 3¹⁸ + 3¹⁹)

2S= 3²⁰-1

S= 3²⁰-1/2

Answer 2

Có : 3S = 3+3^2+....+3^20

2S = 3S - S = (3+3^2+....+3^20)-(1+3+3^2+....+3^19) = 3^20-1

=> S = (3^20-1)/2 = 1743392200

Tk mk nha

Answer 3

3S = 3(1+3+3^2+...+3^19) 3S = 3+3^2+3^3+...+3^20 3S-S = (3+3^2+3^3+...+3^20)-(1+3+3^2+...+3^19) 2S = 3^20-1 S = (3^20-1):2 Ban rui nen ko giai hết da day du cho lam

Answer 4

\(S=1+3+3^2+...+3^{19}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{20}\)

Sau đó cau tính 3S - S la OK

Answer 5

S = 1+3 +3^2 +...+3^18+3^19

Suy ra : 3S = 3 +3^2 +3^3 +...+3^19+3^20

Suy ra 3S - S = ( 3 +3^2 +3^3 +...+3^19+3^20 ) -  (1+3 +3^2 +...+3^18+3^19)

2S = 3^20 - 1

S = 3^20 - 1 /2

Vậy S = 3^20 - 1 /2

P/s : 3^.. = 3 mũ ... nha bạn Vì mình ko biết ghi mũ ^^  

Answer 6

 ( 3²⁰ - 1 ) / 2

  

Answer 7

S = 1 + 3 + 3² + ... + 3¹⁸ + 3¹⁹

\(\Rightarrow\)    S = 3⁰ + 3¹ + 3² + ... + 3¹⁸ + 3¹⁹

\(\Rightarrow\) 3S = 3¹ + 3² + 3³ + 3⁴ + ... + 3¹⁷ + 3¹⁸ + 3¹⁹ + 3 ²⁰

\(\Rightarrow\) 3S - S = 3²⁰ - 3⁰

\(\Rightarrow\) 2S = 3²⁰ - 1 

\(\Rightarrow\) S = \(\frac{3^{20}-1}{2}\)