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

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

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

Từ (1) và (2) 

=> A = 4B

tui ko bít =))))

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)

=>  \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)

=>  \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)

=>  \(S=1-\frac{1}{2^{20}}\)

muộn mất rồi nhưng dù sao cũng cảm ơn bạn

\(\Rightarrow\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

\(\Rightarrow\) \(\frac{1}{2}A=A-\frac{1}{2}=\frac{1}{2^{10}}-\frac{1}{2}\)

Vậy \(A=\left(\frac{1}{2^{10}}-\frac{1}{2}\right):\frac{1}{2}=\frac{2}{2^{10}}-1\)

Do đó \(A+\frac{1}{2^{10}}=\frac{2}{2^{10}}-1+\frac{2}{10}=1\)

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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

A = 2¹ + 2² + ... + 2²⁰¹⁰

A.2 = 2² + 2³ +... + 2²⁰¹¹

A.2 - A = ( 2² +  2³+ ... + 2²⁰¹¹)- ( 2 + 2² + ... + 2²⁰¹⁰ )

A = 2²⁰¹¹  - 2

**** cong chua xuka

giải jup mik mai mik đi học