2¹ + 2² + 2³ + ... + 2¹⁰⁰

Ta có : S = 2 + 2² + 2³ + ... + 2¹⁰⁰

          2S = 2.(2 +  2² + 2³ + ... + 2¹⁰⁰)

          2S = 2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹

        2S - S = (2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹) - (2 +  2² + 2³ + ... + 2¹⁰⁰)

           S = 2¹⁰¹ - 2

\(2^1+2^2+2^3+...+2^{100}\)

Ta có : \(S=2+2^2+2^3+....2^{100}\)

         : \(2S=2.\left(2+2^2+2^3+....+2^{100}\right)\)

        : \(2S=2^2+2^3+.....+2^{100}+2^{101}\)

        : \(2S-S=\left(2^2+2^3+....+2^{100}+2^{101}\right)\)\(-\left(2+2^2+2^3+.....+2^{100}\right)\)

        : \(S=2^{101}-2\)

giả sử A=2^1+2^2+2^3+...2^100(1)

Nhân 2 cả 2 vế ta có:

2A=2^2+2^3+2^4...2^101(2)

Lấy (2)-(1) ta có:

2A-A=(2^2+2^3+2^4...2^101)-(2^1+2^2+2^3+...2^100)

=>A=2^101-2^1

`2/5+1/4`

`=8/20+5/20`

`=13/20`

__

`1/3+3/5`

`=5/15+9/15`

`=14/15`

__

`1/2+1/4`

`=2/4+1/4`

`=3/4`

__

`1/8+5/6`

`=6/48+40/48`

`=46/48`

`=23/24`

2/5 + 1/4
= 8/20 + 5/20
= 13/20
1/3 + 3/5
= 5/15 + 9/15
= 14/15
1/2 + 1/4
= 2/4 + 1/4
= 3/4
1/8 + 5/6
= 6/48 + 40/48
= 46/48
= 23/24

=? như bạn giải rồi đó :)

KO CÓ QUI LUẬT À?

=========================

==========================ko bt 

(  ;-;   )

=A*(1/100-1/10^2)

=0

Bài 1: 

a: \(2A=2^{101}+2^{100}+...+2^2+2\)

\(\Leftrightarrow A=2^{100}-1\)

b: \(3B=3^{101}+3^{100}+...+3^2+3\)

\(\Leftrightarrow2B=3^{100}-1\)

hay \(B=\dfrac{3^{100}-1}{2}\)

c: \(4C=4^{101}+4^{100}+...+4^2+4\)

\(\Leftrightarrow3C=4^{101}-1\)

hay \(C=\dfrac{4^{101}-1}{3}\)

\(=\left(\dfrac{1}{100}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{10^2}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=0\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)=0\)