\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(A=2A-A=2-\frac{1}{2^{2011}}=\frac{2^{2012}-1}{2^{2011}}\)

Nhầm

\(A=2A-A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
 

A=(ghi lại biieur thức)

2A=2+1+1/2+1/2^2+….+1/2^2011

2A-A=A=(2+1+1/2+1/2^2+….+1/2^2011)-(1+1/2+1/2^2+...+1/2^2012)

A=2-1/2^2012

1/2 A= 1/2+1/2^2+1/2^3+1/2^4+...........+1/2^2013

=>A-1/2A= 1 -1/2^2013

=>1/2A=1 -1/2^2013

=>A=(1 - 1/2^2013) : 1/2

I don't know .

2A = 2 + 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹¹

mà A = 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹²

2A - A = 2 - 1/2²⁰¹²

A = 2 - 1/2²⁰¹²

Ta có A=1+1/2+1/2^2+1/2^3+........+1/2^2012

=>2A=2+1+1/2+1/2^2+.......+1/2^2011

=>2A-A=(2+1+1/2+1/2^2+.....+1/2^2011)-(1+1/2+1+1/2^2+1/2^3+.....+1/2^2012)

=>A=\(2-\frac{1}{2^{2012}}\)

\(A=\frac{2^{2013-1}}{2^{2012}}\)

A=đã cho.

1/2*A=1/2+1/2^2+1/2^3+...+1/2^2012+1/2^2013.

A-1/2*A=1-1/2^2013(khử).

1/2*A=1-1/2^2013.

A=2*(1-1/2^2013).

A=2-2/2^2013.

A=2-1/2^2012.

2A=2+1+1/2+1/2^2+1/2^3+...+1/2^2011
2A-A=(2+1+1/2+1/2^2+1/2^3+...+1/2^2011)-(1+1/2+1/2^2+1/2^3+...+1/2^2012)
A=2-2/2012
k cho mik nhé

1+1/2012

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}\)

\(\frac{1}{2}A-A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(-\frac{1}{2}A=\frac{1}{2^{2013}}-1\)

\(A=\frac{\frac{1}{2^{2013}}-1}{2}\)

A = 1+1/2+1/2^2+1/2^3+.....+1/2^2012

2A= 2. (1+1/2+1/22+1/23+.....+1/22012)

2A= 2 + 1 + 1/2 + 1/22 + 1/23 + ...+ 1/22011

2A - A= (2 + 1 + 1/2 + 1/22 + 1/23+ ...+ 1/22011) - (1+1/2+1/22+1/23+.....+1/22012)

1A= 2 + 1 + 1/2 + 1/22 + 1/23 + ...+ 1/22011 - 1-1/2-1/22+1/23+.....+1/22012

1A= 2 - 1/22012

A= 2-1/22012

A= 2 - 1/22012

số mũ nữa nha

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

A=2- 1/2^2012

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

Dat B=1/2+1/2^2+1/2^3+.............+1/2^2012=>2B=1+1/2+1/2^2+.......+1/2^2011

=>2B-B=1/2^2011-1/2^2012=2^2012-2^2011/2^4023

=>A=1+2^2012-2^2011/2^4023

 \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2012}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{2012}}\)

\(\Rightarrow A=\frac{2^{2013}}{2^{2012}}-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)

Vậy \(A=\frac{2^{2013}-1}{2^{2012}}\)

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(2A-A=2-\frac{1}{2^{2012}}\Rightarrow A=2-\frac{1}{2^{2012}}\)

\(A=\frac{2^{2013}}{2^{2012}}-\frac{1}{2^{2012}}=\frac{2^{2012}+1}{2^{2012}}\)

À bạn Yến Nhi, tại sao mà 2²⁰¹³ - 1 lai bằng 2²⁰¹² + 1 thế ?

\(\frac{con}{\frac{chịu}{\frac{thưa}{\frac{cụ}{hi}}\frac{câc}{ạ}}}\)