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

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

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

Từ (1) và (2) 

=> A = 4B

bn ghi thíu đề hay sao ak

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

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

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

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

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

A= 4+2.2+2.2.2+2.2.2.2+.......+{2.2.2.2.2.....} có 20 thừa số 2 

Có số số hạng ở trong khoảng số 2 là:

(20-2)+1=19(số)

Có 20 thừa số 2 suy ra:20.2=40

Tổng là:

(40+2)*19:2=399

A=4+399

A=403

**** nhé Hương Linh xinh xắn

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a = 2 + 2² + 2³ + ... + 2⁶⁰

2a = 2 . 2 + 2² . 2 + 2³ . 2  + ... + 2⁶⁰ . 2

2a =  2² + 2³ + 2⁴ +  ... + 2⁶¹

2a - a = 2⁶¹ - 2

a = 2⁶¹ - 2

Vậy : a = 2⁶¹ - 2

2¹⁰⁰  -2⁹⁹=2¹

2⁹⁸-2⁹⁷=2¹

=> từ 2¹ -> 2¹⁰⁰ có 50 số 2¹

=>2¹⁰⁰-2⁹⁹-2⁹⁸-...-2¹=2¹.50=2⁵⁰

2¹⁰⁰ - 2⁹⁹ = 2² - 2¹