a: 4A=4+4^2+...+4^9

=>3A=4^9-1

=>A=(4^9-1)/3

b: 2A=1+1/2+...+1/2^7

=>A=1-1/256=255/256

c: =1-1/5+1/5-1/9+...+1/85-1/89

=1-1/89=88/89

d: =1/3(3/1*4+3/4*7+...+3/304*307)

=1/3(1-1/4+1/4-1/7+...+1/304-1/307)

=1/3*306/307=102/307

e: E=1-1/2+1/2-1/3+...+1/11-1/12

=1-1/12=11/12

g: =2/5(1-1/6+1/6-1/11+...+1/96-1/101)

=2/5*100/101=40/101

ahihi đồ ngốc,thật sự thì mình không biết nữa...

\(16+64+256+1024+...+16384+65536\)

\(=4^2+4^3+4^4+....+4^7+4^8\)

Đặt : \(A=4^2+4^3+4^4+....+4^7+4^8\)

\(\Rightarrow4A=4^3+4^4+4^5+...+4^8+4^9\)

\(\Rightarrow4A-A=\left(4^3+4^4+4^5+...+4^8+4^9\right)-\left(4^2+4^3+4^4+...+4^7+4^8\right)\)

\(\Rightarrow3A=4^9-4^2\)

\(\Rightarrow A=\frac{4^9-4^2}{3}=87376\)

Vậy : \(16+64+256+1024+...+16384+65536=87376\)

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...+\frac{1}{16384}\)

\(A=\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{14}}\)

\(2^2A=1+\frac{1}{2^2}+...+\frac{1}{2^{12}}\)

\(4A-A=\left(1+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{14}}\right)\)

\(3A=1-\frac{1}{2^{14}}\)

\(A=\frac{1-\frac{1}{2^{14}}}{3}\)

mình cũng đang phân vân câu này

minh moi hoc lop 6

      A =          \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)

     2A =   1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)

2A - A =   1 - \(\dfrac{1}{256}\)

       A   = \(\dfrac{255}{256}\)