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

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

\(2B-B=\left(2^2+2^3+...+2^{11}\right)-\left(2+2^2+...+2^{10}\right)\)

\(B=2^{11}-2\)

a)

Từ 1 đến 23 gồm có (29-1) : 2 + 1 = 15 số

15 số gồm có 7.5 cặp

Mà từng cặp số có tổng bằng 30 (tính từng cặp số ở hai đầu)

Vậy S = 30 * 7.5

          = 225

          =

bạn có thể trình bày theo cách của tiểu học đuọc ko

2A=1+1/2+1/4+1/8+.........+1/512

2A‐A=﴾1+1/2+1/4+1/8+....+1/512﴿‐﴾1/2+1/4+1/8+.....+1/1024﴿

A=1‐1/1024 =1023/1024

vậy A=1023/1024

Đặt A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{1024}\) (1)

Ta có:  2A=\(2+1+\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{512}\) (2)

Từ (1) và (2) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)

\(\Rightarrow A=2-\frac{1}{1024}\)

\(\Rightarrow A=\frac{2047}{1024}\)

\(\text{Đặt }A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

=> \(\frac{1}{2}A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

=> \(\frac{1}{2}A-A=\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

=> \(-\frac{1}{2}A=\frac{1}{2^{11}}-1=\frac{1-2048}{2048}=-\frac{2047}{2048}\)

=> \(A=-\frac{2047}{2048}:\left(-\frac{1}{2}\right)=\frac{2047}{1024}\)

ko bít

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(\Rightarrow A=1-\frac{1}{1024}=\frac{1023}{1024}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{1024}\)

=> \(2A=1+\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{512}\)

=> \(2A-A=1-\frac{1}{1024}\)

=> \(A=\frac{1023}{1024}\)

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

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

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}\)

Đặt:

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

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

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

\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}\)

\(A=1-\dfrac{1}{1024}=\dfrac{2023}{2024}\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

\(\Rightarrow A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{10}}\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}\)

\(\Rightarrow A=1-\dfrac{1}{1024}\)

\(\Rightarrow A=\dfrac{1023}{1024}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}=\dfrac{1023}{1024}\)

\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)

B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)

    = \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)

    = \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)

     = \(\frac{-22}{105}\)

C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

    = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

     = \(1-\frac{1}{7}=\frac{6}{7}\)

\(B=\frac{59}{15}-\frac{29}{7}\)

 \(=\frac{-22}{105}\) ( bài B này mk hk pít giải nhanh mk bấm máy tính lụi)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)\(\left(vì\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\right)\)

\(C=1-\frac{1}{7}=\frac{6}{7}\)

Đặt \(B=4^{2007}+4^{2006}+...+4^2+4+1\)

\(4B=4^{2008}+4^{2007}+...+4^3+4^2+4\)

\(3B=4B-B=4^{2008}-1\Rightarrow B=\frac{4^{2008}-1}{3}\)

\(A=75.\frac{4^{2008}-1}{3}+25=25.\left(4^{2008}-1\right)+25=25.4^{2008}=100.4^{2007}\) Chia hết cho 100

a ) 13/20

B)

C..........................................................

minh dang tính

lấy máy tính mà bấm

trả lời kiểu gì vậy hả bạn lê thế trung

chịu rùi bn ơi

khó quá à

mk ko biết làm

chúc bn học giỏi!

nhae$

=2/6+2/12+2/20+...+2/600

=2(1/(2.3)+1/(3.4)+1/(4.5)+...+1/(24.25)

=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/24-1/25)

=2(1/2-1/25)

=2(25/50-2/50)

=2.23/50

=23/25