\(S_1=1+2+2^2+2^3+..+2^{63}\\ \Rightarrow2S_1=2+2^2+2^3+2^4+...+2^{64}\\ \Rightarrow S_1-2S_1=1-2^{64}\\ \Rightarrow-S_1=1-2^{64}\\ \Rightarrow S_1=2^{64}-1.\)

- Ta có: S₁ = 1 + 2 + 2² + 2³ + … + 2⁶³ = 1 + 2(1 + 2 + 2² + 2³ + … + 2⁶²) (1)

= 1 + 2(S₁ - 2⁶³) = 1 + 2S₁ - 2⁶⁴ S₁ = 2⁶⁴ - 1

H2.right

`#3107.101107`

`S_1 = 1 + 2 + 2^2 + 2^3 + ... + 2^63`

`2S_1 = 2 + 2^2 + 2^3 + .... + 2^64`

`2S_1 - S_1 = (2 + 2^2 + 2^3 + ... + 2^64) - (1 + 2 + 2^2 + 2^3 + ... + 2^63)`

`S_1 = 2 + 2^2 + 2^3 + ... + 2^64 - 1 - 2 - 2^2 - 2^3 - ... - 2^63`

`S_1 = 2^64 - 1`

Vậy, `S_1 = 2^64 - 1.`

A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴

2A-A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴-1+2+2²+2³+...+2⁶²+2⁶³

A=2⁶⁴-1

\(A=1+2+2^2+2^3+..+2^{62}+2^{63}\)

\(2A=2+2^2+2^3+...+2^{63}+2^{64}\)

\(2A-A=2^{64}-1\)

\(A=2^{64}-1\)

A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2(1+2+2²+2³+...+2⁶²+2⁶³)

2A=2+2³+2⁴+2⁵...+2⁶³+2⁶⁴

2A-A=(2+2³+2⁴+2⁵...+2⁶³+2⁶⁴)-(1+2+2²+2³+...+2⁶²+2⁶³)

A=2⁶⁴-1

Nha bạn. Chúc bn ht

A = 1 + 2 + 2² + 2³ + ... + 2⁶² + 2⁶³

2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶³ + 2⁶⁴

A = 2⁶⁴ - 1

tính như nào nhỉ

`1+2+2^2+2^3+....+2^63`

`=2+2+2^2+2^3+....+2^63-1`

`=2.2+2^2+2^3+....+2^63-1`

`=2^2+2^2+2^3+....+2^63-1`

`=2.2^2+2^3+....+2^63-1`

`=2^3+2^3+...2^63-1`

`=2.2^3+....+2^63-1`

`=2^4+....+2^63-1`

`=2^{63}.2-1=2^64-1`

Đặt \(A=1+2+2^2+2^3+...+2^{63}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{64}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{63}\right)=2^{64}-1\left(đpcm\right)\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2017.2019}\)

\(\Leftrightarrow S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(\Rightarrow2S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\right)\)

\(\Leftrightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(\Leftrightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{2}{2019}\)

\(\Leftrightarrow2S=1-\frac{1}{2019}=\frac{2018}{2019}\)

\(\Rightarrow S=\frac{2018}{2019}:2=\frac{1009}{2019}\)

Vậy \(S=\frac{1009}{2019}.\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2017.2019}\)

\(\Rightarrow S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)

\(\Rightarrow S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(\Rightarrow S=\frac{2018}{2019}\)

Ta có : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{2017.2019}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2017.2019}\)

=> \(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2017.2019}\)

=> \(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{2017}-\frac{1}{2019}\)

=> \(2S=1-\frac{1}{2019}=\frac{2018}{2019}\)

=> \(S=\frac{2018}{2019}.\frac{1}{2}=\frac{1009}{2019}\)

S=1+(-2)+3+(-4)+...+99+(-100)=(-2+1)+(-4+3)+...+(-100+99)=(-1)+(-1)+...+(-1)=(-1)*50=-50