\(A=2^{100}-2^{99}-...-2^2-2\)

\(2A=2^{101}-2^{100}-...-2^3-2^2\)

\(2A-A=2^{101}-2^{100}-...-2^3-2^2-2^{100}+2^{99}+...2^2+2\)

\(A=2^{101}-\left(2^{100}-2^{100}+2^{99}-2^{99}+...+2^2-2^2+-2\right)\)

\(A=2^{101}+2\)

2¹⁰⁰  -2⁹⁹=2¹

2⁹⁸-2⁹⁷=2¹

=> từ 2¹ -> 2¹⁰⁰ có 50 số 2¹

=>2¹⁰⁰-2⁹⁹-2⁹⁸-...-2¹=2¹.50=2⁵⁰

\(B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{100^2}{99.101}=\dfrac{2.3.4...100}{1.2.3...99}.\dfrac{2.3.4..100}{3.4.5...101}=100.\dfrac{2}{101}=\dfrac{200}{101}\)

2¹⁰⁰ - 2⁹⁹ = 2² - 2¹

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

Lời giải:

$S=5(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023})$

$=5(\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2023-2022}{2022.2023})$

$=5(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2022}-\frac{1}{2023})$

$=5(\frac{1}{2}-\frac{1}{2023})=\frac{10105}{4046}$