Question

So sánh A,B

A=\(2^{2020}-1\)

B=1+2+\(2^3+2^3+....+2^{2019}\)

Answer 1

Sửa đề : \(B=1+2+2^2+2^3+...+2^{2019}\)

\(2B=2+2^2+2^3+2^4...+2^{2019}+2^{2020}\)

\(2B-B=\left(2+2^2+2^3+...+2^{2020}\right)-\left(1+2+2^2+...+2^{2019}\right)\)

\(\Leftrightarrow2B-B=B=2^{2020}-1=A\Rightarrow B=A\)

Answer 2

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Answer 3

Đây hình như là Toán lớp 6 mà! B = 1 + 2 + 2² + 2³ +  ... + 2²⁰¹⁹ mới đúng chứ nhỉ?! :>>

Ta có: \(B=1+2+2^2+...+2^{2019}\)

\(\Rightarrow2B=2+2^2+2^3+...+2^{2020}\)

\(2B-B=\left(2+2^2+2^3+...+2^{2020}\right)-\left(1+2+2^2+...+2^{2019}\right)\)

\(B=2^{2020}-1\)

\(\Rightarrow A=B\)

Vậy \(A=B\).

Answer 4

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Answer 5

B = 1 + 2 + 2² + 2³ + ... + 2²⁰¹⁹

2B = 2( 1 + 2 + 2² + 2³ + ... + 2²⁰¹⁹ )

2B = 2 + 2² + 2³ + ... + 2²⁰²⁰

B = 2B - B

    = 2 + 2² + 2³ + ... + 2²⁰²⁰ - ( 1 + 2 + 2² + 2³ + ... + 2²⁰¹⁹ )

    = 2 + 2² + 2³ + ... + 2²⁰²⁰ - 1 - 2 - 2² - 2³ - ... - 2²⁰¹⁹

    = 2²⁰²⁰ - 1

=> A = B