Question

Tính tổng:

\(A=2^0+2^1+2^2+...+2^{2010}\)

Answer 1

\(2A=2+2^2+2^3+...+2^{2011}\)

\(A=2^0+2^1+2^2+...+2^{2010}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)

\(\Rightarrow A=2^{2011}-1\)

Answer 2

Tính tổng:

$A=2^0+2^1+2^2+...+2^{2010}$A=20+21+22+...+22010

Câu hỏi tương tự Đọc thêmToán lớp 6              

Answer 3

Nguyễn Đắc Phúc An Thằng này bị dean à

Answer 4

=>A=1+2¹+2²+...2²⁰¹⁰

=>2A=2(1+2¹+2²+...2²⁰¹⁰)

=>2A=2.1+2.2¹+2.2²+...+2.2²⁰¹⁰

=>2A=2+2²+2³+...+2²⁰¹¹

=>2A-A=(2+2²+2³+...+2²⁰¹¹)-(1+2¹+2²+...2²⁰¹⁰)

=>A=2²⁰¹¹-1

Answer 5

Ta có:

=>A=1+2¹+2²+...2²⁰¹⁰

=>2A=2(1+2¹+2²+...2²⁰¹⁰)

=>2A=2.1+2.2¹+2.2²+...+2.2²⁰¹⁰

2A=2+2²+2³+...+2²⁰¹¹

2A-A=(2+2²+2³+...+2²⁰¹¹)-(1+2¹+2²+...2²⁰¹⁰)

=>A=2²⁰¹¹-1

Vậy......