moi nguoi oi help me

S = 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 9

2S = 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + .... + 2 ^ 10

2S - S = ( 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + .... + 2 ^ 10 )

           -  ( 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 9 )

S        = 2 ^ 10 - 1

S       = 2 ^ 8 . 2 ^ 2 - 1

S       = 2 ^ 8 . 4 - 1

S       < 2 ^ 8 . 1 < 5 . 2 ^ 8

Vậy S < 5 . 2 ^ 8

\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2015}{5^{2014}}\Rightarrow4S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}-\frac{2015}{5^{2015}}\)

Đặt B = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\)

 => 5B = \(5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)

=> 4B = \(5-\frac{1}{5^{2014}}

S>5 . 2⁹⁸

A = 1 + 2 + 2² + 2³ + .... +2⁹⁸

2A = 2(1 + 2 + 2² + 2³ +....+2⁹⁸)

2A = 2 + 2² + 2³ + 2⁴ +....+2⁹⁹

2A - A = (2 + 2² + 2³ + 2⁴ +.....+ 2⁹⁹) - (1 + 2 + 2² + 2³ +.....+2⁹⁸)

A = ( 2 - 2 ) +  ( 2² - 2² ) + (2³ - 2³).....+(2⁹⁸ - 2⁹⁸) + 2⁹⁹ - 1

A = 0 + 0 + 0 +.....+0  + 2⁹⁹ - 1

A = 2⁹⁹ - 1

So sánh : 

2⁹⁹ - 1 và 5 . 2⁹⁸

2⁹⁹ - 1 < 5 . 2⁹⁸

( Biết hay sai thì chịu nhe hehe phần so sánh tui 

Bài 1: 

a: \(2P=2^{101}-2^{100}+2^{98}-2^{97}+...+2^3-2^2\)

=>\(3P=2^{101}-2\)

hay \(P=\dfrac{2^{101}-2}{3}\)

b: \(5Q=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2+5\)

=>\(6Q=5^{101}+1\)

hay \(Q=\dfrac{5^{101}+1}{6}\)

Bài 2:

Ta có: \(5^x.5^{x+2}\le10^{18}\div2^8\)

\(\Rightarrow5^{x+x+2}\le\left(10\div2\right)^{18}\)

\(\Rightarrow5^{2x+2}\le5^{18}\)

\(\Rightarrow2x+2\le18\Rightarrow2x\le16\Rightarrow x\le8\)

\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

Bài 3:

Ta có: \(S=1+2+2^2+...+2^9=\left(2+2^2+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)

\(=2^{10}-1\left(1\right)\)

Ta có: \(5\times2^8=\left(2^2+1\right)\times2^8=2^{10}+2^8\left(2\right)\)

Từ (1) và (2) \(\Rightarrow S< 5\times2^8\)