Question

cho n = 1 + 3 + 3² + 3³ + .... + 3⁹⁹

tìm 2n + 1

Answer 1

Ta có:

n = 1 + 3 + 3² + 3³ + ... + 3⁹⁹

=> 3n = 3 (  1 + 3 + 3² + 3³ + ... + 3⁹⁹ )

3n = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰

3n - n = 2n = 3¹⁰⁰ - 1

=> 2n + 1 = 3¹⁰⁰

Answer 2

n = 1 + 3 + 3² + 3³ + ... + 3⁹⁹

3n = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰

3n - n = (3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰) - (1 + 3 + 3² + 3³ + ... + 3⁹⁹)

2n = 3¹⁰⁰ - 1

2n + 1 = 3¹⁰⁰

Ủng hộ mk nha ^_-

Answer 3

\(n=1+3+3^2+3^3+...+3^{99}\)

\(=>3n=3+3^2+3^3+3^4+....+3^{100}\)

\(=>3n-n=\left(3+3^2+3^3+3^4+...+3^{100}\right)-\left(1+3+3^2+3^3+....+3^{99}\right)\)

\(=>2n=3^{100}-1=>2n+1=3^{100}-1+1=3^{100}\)

Answer 4

n=1+3+3²+3³+...+3⁹⁹

3n=3+3²+3³+3⁴+....+3¹⁰⁰

2n=3n-n=3¹⁰⁰-1(sau khi đã khử những số đối nhau nha,bước này chị làm tắt)

2n+1=3¹⁰⁰-1+1

=3¹⁰⁰

Vậy 2n+1=3¹⁰⁰

Chúc em học tốt^^

Answer 5

3n = 3 . ( 1 + 3 + 3² + 3³ + ..... + 3⁹⁸ + 3⁹⁹ )

3n = 3 + 3² + 3³ + 3⁴ + ...... + 3⁹⁹ + 3¹⁰⁰

3n + 1 = 1 + 3 + 3² + 3³ + 3⁴ + ...... + 3⁹⁹ + 3¹⁰⁰

=> 3n + 1 = n + 3¹⁰⁰

=> 3n - n + 1 = 3¹⁰⁰

=> 2n + 1 = 3¹⁰⁰

 

Answer 6

Ta có: \(N=1+3+3^2+...+3^{99}\)

\(\Rightarrow3N=3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3N-N=\left(3+3^2+...+3^{100}\right)-\left(1+3+3^2+....+3^{99}\right)\)

\(\Rightarrow2N=3^{100}-1\)

\(\Rightarrow2^n+1=3^{100}\)

Answer 7

N = 1 + 3 + 3² + 3³ + ... + 3⁹⁹

=>3N = 3.(1+3+3²+3³+...+3⁹⁹)

=>3N = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰

=>3N - N = (3 + 3² + 3³ + 3⁴ + ... + 3¹⁰⁰) - (1 + 3 + 3² + 3³ + ... + 3⁹⁹)

=>2N = 3¹⁰⁰ - 1

=>2n + 1 = 3¹⁰⁰

Answer 8

Ta có:n = 1 + 3 + 3² + 3³ + ... + 3⁹⁹

=> 3n = 3 + 3² + 3³ + ... + 3¹⁰⁰

=> 3n - n = 3¹⁰⁰ - 1

=> 2n = 3¹⁰⁰ - 1

\(=>2n+1=3^{100}\)