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

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

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

hay \(D=\dfrac{3^{151}-3^{100}}{2}\)

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

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

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

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

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

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

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           

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

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

2A = 3¹⁰¹ - 1

A = \(\frac{3^{101}-1}{2}\)

3B = 4B - B = (4 + 4² + ... + 4⁵¹) - (1 + 4 + 4² + ... + 4⁵⁰)

3B = 4⁵¹ - 1

B = \(\frac{4^{51}-1}{3}\)

2A = 3A - A = ( 3 + 3^{2  +}  3³  +  ...  +  3¹⁰¹ )  - ( 1 + 3 + 3² +  3³  +  ... + 3¹⁰⁰ ) 

2A = 3¹⁰¹ - 1

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

3B =  4B - B = ( 4 + 4²  + ... +  4⁵¹) - ( 1 + 4 + 4² +...+ 4⁵⁰ )

3B = 4⁵¹ - 1

B = 4⁵¹ - 1 : 3

A=1+3+3²+...+3¹⁰⁰

3A=3+3²+3³+...+3¹⁰¹

=>3A+1=1+3+3²+...+3¹⁰⁰+3¹⁰¹=A+3¹⁰¹

=>3A-A=3¹⁰¹-1

2A=3¹⁰¹-1

A=(3¹⁰¹-1)/2

B=1+4+4²+...+4⁵⁰

4B=4+4²+...+4⁵¹

4B+1=1+4+4²+...+4⁵⁰+4⁵¹=B+4⁵¹

=>4B-B=4⁵¹-1

3B=4⁵¹-1

B=(4⁵¹-1)/3

Đầu luỹ thừa đuôi số hạng 

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

2A = 3¹⁰¹ - 1

A = \(\frac{3^{101}-1}{2}\)

3B = 4B - B = (4 + 4² + ... + 4⁵¹) - (1 + 4 + 4² + ... + 4⁵⁰)

3B = 4⁵¹ - 1

B = \(\frac{4^{51}-1}{3}\)

A=1+3+3²+...+3¹⁰⁰

3A=3+3²+3³+...+3¹⁰¹

=>3A+1=1+3+3²+...+3¹⁰⁰+3¹⁰¹=A+3¹⁰¹

=>3A-A=3¹⁰¹-1

2A=3¹⁰¹-1

A=(3¹⁰¹-1)/2

B=1+4+4²+...+4⁵⁰

4B=4+4²+...+4⁵¹

4B+1=1+4+4²+...+4⁵⁰+4⁵¹=B+4⁵¹

=>4B-B=4⁵¹-1

3B=4⁵¹-1

B=(4⁵¹-1)/3

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:

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

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

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

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

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