B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{2024}\) + 2\(^{2026}\)

2\(^2\) B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\)

4B - B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\) - 1 - 2\(^2\) - 2\(^4\) - ... - 2\(^{2024}\) - 2\(^{2026}\)

3B = (2\(^2\) - 2\(^2\)) + (2\(^4\) - 2\(^4\)) +...+ (2\(^{2026}\) - \(2^{2026}\)) + (2\(^{2028}\) - 1)

3B = 0 + 0 +... + 0 + 2\(^{2028}\) - 1

3B = 2\(^{2028}\) - 1

B = \(\frac{2^{2028}-1}{3}\)

Ta có: \(B=1+2^2+2^4+\cdots+2^{2024}+2^{2026}\)

=>\(4B=2^2+2^4+2^6+\ldots+2^{2026}+2^{2028}\)

=>\(4B-B=2^2+2^4+2^6+\cdots+2^{2026}+2^{2028}-1-2^2-\cdots-2^{2026}\)

=>\(3B=2^{2028}-1\)

=>\(B=\frac{2^{2028}-1}{3}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2024}}\\ =>2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2023}}\\ =>2A-A=A=1-\dfrac{1}{2^{2024}}=\dfrac{2^{2024}-1}{2^{2024}}\)

e cảm ơn

Đặt \(A=1+2+2^2+2^3+\cdots+2^{2023}\)

=>\(2A=2+2^2+2^3+2^4+\cdots+2^{2024}\)

=>\(2A-A=2+2^2+2^3+2^4+\cdots+2^{2024}-1-2-\cdots-2^{2023}\)

=>\(A=2^{2024}-1\)

Ta có: \(2^{2024}-\left(1+2+2^2+2^3+\cdots+2^{2023}\right)\)

\(=2^{2024}-\left(2^{2024}-1\right)=1\)

2^ 2024 -( 1 + 2 + 2 ^ 2 + 2 ^ 3 +...+2^ 2023 )

2^2024 -(2^2024-1)

1

Nhưng dài lắm chắc lớp 6 ko hiểu đc nên tui chỉ viết ngắn gọn

BẠN THÔNG CẢM

ban tra loi di

1+1-1+1=

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{15}\left(1+2+...+15\right)+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(A=1+\frac{1}{2}\cdot3+\frac{1}{3}\cdot6+\frac{1}{4}\cdot10+...+\frac{1}{15}+\left[\frac{\left(1+15\right)\cdot15}{2}\right]+\frac{1}{16}\cdot\left[\frac{\left(16+1\right).16}{2}\right]\)

\(A=1+\frac{3}{2}+2+\frac{5}{2}+....+\frac{1}{15}\cdot120+\frac{1}{16}\cdot136\)

\(A=1+\frac{3}{2}+2+\frac{5}{2}+...+8+\frac{17}{2}\)

\(A=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)\)

Đặt \(B=1+2+...+8\)

      \(C=\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(B=1+2+...+8\)

\(\text{Ta thấy tổng B là dãy các số hạng liên tiếp từ 1 đến 8 }\)

\(\Rightarrow\text{số số hạng của B là}:\)\(\left(8-1\right)\div1+1=8\left(sh\right)\)

                    \(\text{Tổng B là }:\)\(\frac{\left(1+8\right)\cdot8}{2}=36\)

\(C=\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(\Rightarrow C=\frac{3+5+...+17}{2}\)

Đặt \(D=3+5+...+17\)

\(\text{số số hạng của D là}:\)\(\left(17-3\right)\div2+1=8\left(sh\right)\)

               \(\text{Tổng D là }:\)\(\frac{\left(3+17\right)\cdot8}{2}=80\)

\(\Rightarrow C=\frac{80}{2}=40\)

Thay B và C vào biểu thức A , ta được 

\(A=36+40=76\)

Vậy A = 76 

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)\)\(+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(\Rightarrow A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)

\(\Rightarrow A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(\Rightarrow A=\frac{\frac{17.18}{2}-1}{2}=76.\)

Vậy \(A=76.\)

1/A=1.2¹.2².2³.2⁴.25                                                               câu 2 làm tương tự                                                            

A.2=2.2².2³.2⁴.2⁵.2⁶                                

A.2-A=(2.2².2³.2⁴.2⁵.2 mũ 6)-(1.2¹.2².2³.2⁴.2⁵)

A=2⁶-1

3 A=1+3+3²+3³+...3⁷

3.A=3+3²+3³+3⁴...+3⁸

2A=3⁸-1

A=(3⁸-1):2

1/A=1.2¹.2².2³.2⁴.25   

A.2=2.2².2³.2⁴.2⁵.2⁶                                

A.2-A=(2.2².2³.2⁴.2⁵.2mũ 6

)-(1.2¹.2².2³.2⁴.2⁵)

A=2⁶-1

3 A=1+3+3²+3³+...3⁷

3.A=3+3²+3³+3⁴...+3⁸

2A=3⁸-1

A=(3⁸-1):2

Câu 1 Tính 

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)

Câu 2 Tính 

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)

\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

Câu 3 

a) Ta có : M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹ (1)

=> 3M = 3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰  (2)

Lấy (2) trừ (1) theo vế ta có : 

3M - M = (3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰) - ( M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹)

=>  2M = 3¹²⁰ - 1

=>    M = \(\frac{3^{120}-1}{2}\)

b) M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

        = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

        = (1 + 3 + 3²) + 3³(1 + 3 + 3²) + ... + 3¹¹⁷(1 + 3 + 3²)

        = 13 + 3³.13 + ... + 3¹¹⁷.13

        = 13(1 + 3³ + ... + 3¹¹⁷) \(⋮\)13

=> M \(⋮\)13

M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

= (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + ... + (3¹¹⁶ + 3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

= (1 + 3 + 3² + 3³) + 3⁴(1 + 3 + 3² + 3³) + ... + 3¹¹⁶(1 + 3 + 3² + 3³)

= 40 + 3⁴.40 + ... + 3¹¹⁶.40

= 40(1 + 3⁴ + ... + 3¹¹⁶) 

= 5.8.(1 + 3⁴ + ... + 3¹¹⁶)  \(⋮\)5

4) Tính 

A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1

=> 2A =  2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1

Lấy 2A trừ A theo vế ta có : 

2A - A = (2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1) - (2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1)

=>   A = 2¹⁰¹ - 2¹⁰⁰ - 2¹⁰⁰ + 1

=>   A = 2¹⁰¹ - (2¹⁰⁰ + 2¹⁰⁰) + 1

=>   A  = 2¹⁰¹ - 2¹⁰⁰ . 2 + 1

=>   A = 1

Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

          = 99.100.101 

=> C = 99.100.101 : 3 =  333300

b) Ta có : D = 2² + 4² + 6² + ... + 98²

                    = 2²(1² + 2²  + 3² + ... + 49²

                    =  2² .(1² + 2²  + 3² + ... + 49²)

                    = 2².(1.1 + 2.2 + 3.3 + ... + 49.49)

                    = 2².[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]

                    = 2².[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]

Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50

=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

          = 49.50.51 

=> E = 49.50.51/3 = 41650

Khi đó D = 2².[41650 - (1 + 2 + 3 + 4 + ... + 49)]

               = 2².[41650 - 49(49 + 1)/2]

               = 2².[41650 - 1225 

               = 2².40425

               = 161700

=> D = 161700