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

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

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

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

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

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

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

\(D=3^{100}+3^{101}+...+3^{149}+3^{150}\)

nên \(3D=3^{101}+3^{102}+...+3^{150}+3^{151}\)

\(\Leftrightarrow2\cdot D=3^{151}-3^{100}\)

hay \(D=\dfrac{3^{151}-3^{100}}{2}\)

\(3D=3^{101}+3^{102}+3^{103}+...+3^{150}+3^{151}\\ 3D-D=3^{151}-3^{100}\\ 2D=3^{151}-3^{100}\\ D=\dfrac{3^{151}-3^{100}}{2}\)

Bài 1:

$B=1+3+3^2+3^3+...+3^{100}$

$=1+(3+3^2)+(3^3+3^4)+...+(3^{99}+3^{100})$

$=1+3(1+3)+3^3(1+3)+...+3^{99}(1+3)$

$=1+(1+3)(3+3^3+...+3^{99})=1+4(3+3^3+....+3^{99})$

$\Rightarrow B$ chia 4 dư 1.

Bài 2:

$C=5-5^2+5^3-5^4+...+5^{2023}-5^{2024}$

$5C=5^2-5^3+5^4-5^5+...+5^{2024}-5^{2025}$

$\Rightarrow C+5C=5-5^{2025}$

$6C=5-5^{2025}$

$C=\frac{5-5^{2025}}{6}$

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

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

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

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

___________

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

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

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

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

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

___________

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

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

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

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

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

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

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

a) \(S=1+2+2^2+..+2^{2022}\)

\(2S=2+2^2+2^3+...+2^{2023}\)

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

\(3S=3^2+3^3+...+3^{2023}\)

\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

  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 = \(\dfrac{3^{101}-1}{2}\)

 a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)

b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)

c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)

\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)

\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)

Tương tự câu d,e,f bạn tự làm nhé