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

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^9}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{10}}\right)\)

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

\(\Leftrightarrow A+\dfrac{1}{2^{10}}=1\left(đpcm\right)\)

1+1=3

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

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

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

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

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

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

\(A+\frac{1}{2^{10}}=1\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot2}-\dfrac{1}{2\cdot2\cdot2}+\dfrac{1}{2\cdot2\cdot2}-\dfrac{1}{2\cdot2\cdot2\cdot2}+.....+\dfrac{1}{2^{10}}\)

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

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

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

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

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

       A= 1 - \(\frac{1}{2^{10}}\)

       A= \(\frac{1023}{1024}\)

      một số chỗ hơi tắt bạn thông cảm nha

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B

\(\text{Ta có:}2;6;10;...;8010\text{ đều chia 4 dư 2}\)

\(\Rightarrow X\equiv2^2+3^2+4^2+....+2004^2\left(mod\text{ }10\right)\)

\(\text{ mà:}1^2+2^2+3^2+....+2004^2=\frac{2004.2005.4009}{6}=333.2005.4009\)

\(\Rightarrow X\equiv333.2005.4009-1\left(\text{mod 10}\right)\equiv3.5.9-1\equiv4\left(\text{mod 10}\right)\)

Vậy X có chữ số tận cùng là 4

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

\(< 1+\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{2^2}\right)+..........\left(\frac{1}{2^9}+\frac{1}{2^9}+....+\frac{1}{2^9}\left(\text{512 số hạng }\frac{1}{2^9}\right)\right)\)

\(=1+1+1+1+1+1+1+1+1+1\)

\(=10\left(\text{điều phải chứng minh}\right)\)

\(\text{bài 2 câu b tương tự câu a}\)

câu 2 mình chưa hiểu lắm bạn có thể giải thích cho mình được không ạ? Mình cảm ơn ạ ^^

a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...

b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

Thay B vào A ta được:

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

Vậy....

c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)

\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)

Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)

d, chắc là đề sai

e, giống câu a

a) S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰

   = (2 + 2²) + (2³ + 2⁴) +.....+ (2⁹ + 2¹⁰)

   = 2(1 + 2) + 2³(1 + 2) +....+ 2⁹(1 + 2)

   = 3.(2 + 2³ +.... + 2⁹) chia hết cho 3

   => S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰ chia hết cho 3 (Đpcm)

b) 1+3²+3³+3⁴+...+3⁹⁹

=(1+3+3²+3³)+....+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

=40+.........+3⁹⁶.40

=40.(1+....+3⁹⁶) chia hết cho 40 (đpcm)

ai trả lời giúp mình mình k cho

BÀI 1:

S = 2 + 2² + 2³ + 2⁴ + ..... + 2¹⁰

= (2 + 2²) + ( 2³ + 2⁴) + ..... + (2⁷ + 2⁸) + (2⁹ + 2¹⁰)

= 2(1 + 2) + 2³(1 + 2) + ..... + 2⁷(1 + 2) + 2⁹(1 + 2)

= 3(2 + 2³ + .... + 2⁷ + 2⁹)    \(⋮3\)

BÀI 2:

1 + 3 + 3² + 3³ + ....... + 3⁹⁹

= (1 + 3 + 3² + 3³) + ..... + (3⁹⁶ + 3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= (1 + 3 + 3² + 3³) + ..... + 3⁹⁶(1 + 3 + 3² + 3³)

= 40(1 + 3⁴ + ..... + 3⁹⁶)     \(⋮40\)

ta có

\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};.......;\frac{1}{10^2}<\frac{1}{9.10}\)

=> \(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+......+\frac{1}{9}-\frac{1}{10}\)

    \(A<1-\frac{1}{10}=\frac{9}{10}<1\)

vậy A< 1