giải jup mik mai mik đi học

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

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

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

=>  \(S=1-\frac{1}{2^{20}}\)

muộn mất rồi nhưng dù sao cũng cảm ơn bạn

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

S = 1² + 2² + ... + 10²

S = 1 + 4 + ... + 100

S = 385

2S=2+2²+2³+....+2¹⁰+2¹¹

2S-S=2¹¹-1

S=2^11-1

\(B=1+2+2^2+...+2^6.\)

\(=>4B=2^2+2^3+...+2^8\)\(\left(1\right)\)

\(A=2^2+2^3+...+2^8\)\(\left(2\right)\)

Từ (1) và (2) 

=> A = 4B

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

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

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

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

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

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

tui ko bít =))))

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)

:)))