S = 22 + 42 + 62 + ... + 202

   = (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2

   = 22.12 + 22.22 + 22.32 + ... + 22.102

   = 22 (12 + 22 + ... + 102 )

   = 4 . 385

   = 1540

= (1x2)^2 (2x2)^2 (3x2)^2 (4x2)^2 ..... (9x2)^2 (10x2)^2 
= 1^2 x 2^2 2^2 x 2^2 3^2 x 2^2 4^2 x 2^2 ..... 9^2 x 2^2 10^2 x 2^2 
= (1^2 2^2 3^2 4^2 ..... 9^2 10^2) x 2^2 
= 385 x 2^2 = 385 x 4 = 1540

\(\left[1-\frac{1}{2^2}\right]\left[1-\frac{1}{3^2}\right]\left[1-\frac{1}{4^2}\right]...\left[1-\frac{1}{10^2}\right]\)

\(=\left[1-\frac{1}{4}\right]\left[1-\frac{1}{9}\right]\left[1-\frac{1}{16}\right]...\left[1-\frac{1}{100}\right]\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}\)

Tự tính :v

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)

=>  \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)

=>  \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)

=>  \(S=1-\frac{1}{2^{20}}\)

muộn mất rồi nhưng dù sao cũng cảm ơn bạn

\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)

\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)

\(S=2^2.\dfrac{2022}{2023}\)

\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)

\(10^2\cdot10^3\cdot10^4\cdot10^5\cdot10^6\)

\(=10^{2+3+4+5+6}\)\

\(=10^{20}\)

Trả lời:

10² . 10³ . 10^{4 .} 10⁵ . 10⁶

= 10^2+3+4+5+6

= 10²⁰

Hok tốt!

Đạt A = 2 + 2 ² + ... + 2 ⁵⁰⁰

2A  = 2 ² + 2 ³ + .... + 2 ⁵⁰¹

2A - A = ( 2 ² + 2 ³ + .... + 2 ⁵⁰¹ )

           -  ( 2 + 2 ² + ... + 2 ⁵⁰⁰ )

A         = 2 ⁵⁰¹  - 2

HAY LẮM

Dat A =1+2+2²+2³+...+2⁵⁰      

=> 2A=2+2²+2³+2⁴+...+2⁵¹

=> 2A-A= 2+2²+2³+2⁴+...+2⁵⁰ - ( 1+2+2²+2³+...+2⁵⁰ )

=> A=2⁵¹-1

2²s=2+2²+...+2²⁰²⁰

4S-S=(2+2²+...+2²⁰²⁰⁾-(1+2+2²+....+2²⁰¹⁸)

3S=2²⁰²⁰-1

S=(2²⁰²⁰-1):3

cảm ơn cậu 

\(S=1+2+2^2+...+2^{2018}\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{2019}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+...+2^{2018}\right)\)

\(\Rightarrow S=2^{2019}-1\)

Vậy...(tự kết luận)

:)))