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

\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)

Trừ theo vế:

\(\Rightarrow3A-A=\left(3+3^2+3^3+...3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

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

xin lỗi bài trên của mình làm sai

Ta có: 3A = 3.(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰⁾ 

3A = 3+3²+3³+...+3¹⁰⁰+3¹⁰¹

Suy ra: 3A – A = (3+3²+3³+...+3¹⁰⁰+3¹⁰¹)−(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰)

2A = 3¹⁰¹−1

⇒ A = 3¹⁰¹−1

             2               

Vậy A = 3¹⁰¹−1

                 2           

em den lam

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)

\(A=2^{21}-2\)

___________

\(B=5+5^2+...+5^{50}\)

\(5B=5^2+5^3+...+5^{51}\)

\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)

\(4B=5^{51}-5\)

\(B=\dfrac{5^{51}-5}{4}\)

___________

\(C=1+3+3^2+...+3^{100}\)

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

\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)

\(2C=3^{101}-1\)

\(C=\dfrac{3^{101}-1}{2}\)

2A= 2(2+2²+2³+...+2¹⁹+2²⁰)

2A= 2²+2³+2⁴+...+2²⁰+2²¹

2A-A=(2²+2³+2⁴+...+2²⁰+2²¹)-(2+2²+2³+...+2¹⁹+2²⁰)

A=2²¹-2

Vậy A=2²¹-2

Làm tương tự nhee

khó v

Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12

Mà đoạn 2A sai nhé bạn, sửa lại:

2A = 3101−13101−1 2A=-10001

A=-10001/2

A=-5000,5

Vậy A=-5000,5

\(B=1+3+3^2+3^3+...+3^{100}+3^{101}\)

\(\Rightarrow3B=3+3^2+3^3+3^4+...+3^{101}+3^{102}\)

\(\Rightarrow3B-B=3^{102}-1\)

\(\Leftrightarrow2B=3^{102}-1\)

\(\Leftrightarrow B=\dfrac{3^{102}-1}{2}\)

  A = 1 +  3  + 3² + 3³ + ... + 3¹⁰⁰

3A = 3 + 3² + 3³ +3⁴+ .... + 3¹⁰¹

3A - A = (3 + 3² + 3⁴ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

2A     = 3 + 3² + 3⁴ + ... + 3¹⁰¹ - 1 - 3 - 3² - 3³ - ... - 3¹⁰⁰

2A = (3 - 3) + (3² - 3²) + ... + (3¹⁰⁰ - 3¹⁰⁰) + (3¹⁰¹ - 1)

2A = 3¹⁰¹ - 1

A = \(\dfrac{3^{101}-1}{2}\)

Tham khảo

Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)

3A = 3+32+33+...+3100+31013+32+33+...+3100+3101

Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)

2A = 3101−13101−1

⇒⇒ A = 3101−123101−12

Vậy A = 3101−12

A=3 mũ 101-1 phân số2

\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)