a) A = 2 + 2² + 2³ + ... + 2¹⁰⁰

2A = 2² + 2³ + 2⁴ + ... + 2¹⁰¹

2A - A = (2² + 2³ + 2⁴ + ... + 2¹⁰¹) - (2 + 2² + 2³ + ... + 2¹⁰⁰)

A = 2¹⁰¹ - 2

b) B = 1 + 3 + 3² + ... + 3²⁵⁵

3B = 3 + 3² + 3³ + ... + 3²⁵⁶

3B - B = (3 + 3² + 3³ + ... + 3²⁵⁶) - (1 + 3 + 3² + ... + 3²⁵⁵)

2B = 3²⁵⁶ - 1

B = \(\frac{3^{256}-1}{2}\)

c) C = 1 + 4 + 4² + ... + 4¹⁰⁰

4C = 4 + 4² + 4³ + ... + 4¹⁰¹

4C - C = (4 + 4² + 4³ + ... + 4¹⁰¹) - (1 + 4 + 4² + ... + 4¹⁰⁰)

3C = 4¹⁰¹ - 1

C = \(\frac{4^{101}-1}{3}\)

d) D = 1 + 5 + 5² + ... + 5¹⁰⁰⁰

5D = 5 + 5² + 5³ + ... + 5¹⁰⁰¹

5D - D = (5 + 5² + 5³ + ... + 5¹⁰⁰¹) - (1 + 5 + 5² + ... + 5¹⁰⁰⁰)

4D = 5¹⁰⁰¹ - 1

D = \(\frac{5^{1001}-1}{4}\)

cứ tổng hai số hạng sẽ chia hết cho 3 nhé 

\(A=1+2+2^2+2^3+...+2^{11}\)

\(A=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right)\)

\(A=3+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)\)

\(A=3+2^2.3+...+2^{10}.3\)

\(A=3\left(1+2^2+...+2^{10}\right)\)

\(\Rightarrow A⋮3\)                  

Vậy \(A⋮3\)

  !!!

ok mình làm đc r thank you bạn nha@

A=1+3/2^3+4/2^4+5/2^5+...100/2^100 

1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101 

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3) 

=[1-(1/2)^101]/(1-1/2) -100/2^101 

=(2^101 -1)/2^100 - 100/2^101 

=> A = (2^101 -1)/2^99 - 100/2^100 

Bạn ơi khó hiểu quá bạn giải chi tiết hơn giúp mình nhé mình sẽ k cho bạn 2 cái nhé

\(\frac{B}{A}=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{1+\left[\frac{1}{99}+1\right]+\left[\frac{2}{98}+1\right]+\left[\frac{3}{97}+1\right]+...+\left[\frac{98}{2}+1\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{100\cdot\left[\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right]}{\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}=100\)

Vậy : \(\frac{B}{A}=100\)

Ta có:

\(B=\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}\)

\(=\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1\)

\(=\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}\)

\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

\(=100.A\)

\(\Rightarrow\frac{B}{A}=100\)

a,  2015^2 - 2014^2

=(2015-2014)(2015+2014)

=1.4029

=4029

b,  1^2 - 2^2 + 3^2 - 4^2 + ......+  99^2 - 100^2

=(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100)

=-(1+2)-(3+4)-...-(99+100)

=-1-2-3-4-...-99-100

=(-1-100).100:2=-5050