1. (1+1/2).(1+1/2^2).(1+1/2^3)....(1+1/2^100) < 3
2. 1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+ 1024/(5^1024+1) <1/4
3. 3/(1!+2!+3!)+4/(2!+3!+4!)+...+100/(98!+99!+100!) <1/2
C=3/2+3/4+3/8+3/16+...+3/128
D=1/2+1/4+1/8+...+1/1024
E=5/2+5/8+5/32+5/128+5/512+5/2048
CMR :1/(5+1)+2/(5^2+1)+4/(5^4+1)+⋯+1024/(5^1024+1) <1/4
A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
A x 2 = 1 - 1/1024 + A
A x 2 - A = 1 - 1/1024
A = 1 - 1/1024
A = 1023 /1024
CMR :1/(5+1)+2/(5^2+1)+4/(5^4+1)+⋯+1024/(5^1024+1) <1/4
àm sao để đăng câu hỏi z
chứng minh1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+1024/(5^1024+1)<1/4
Tính:
a) C= 1+1/2+(1/2)^2+(1/2)^3+....+(1/2)^3+...+(1/2)^100.
b) D=(1/2)^10*5-(1/4)^5*3
1/1024*1/3-(1/2)611
Chứng Minh Rằng:
\(\frac{1}{5+1}+\frac{2}{5^2+2}+\frac{4}{5^4+1}+...+\frac{1024}{5^{1024}+1}< \frac{1}{4}\)
a/ 4^(x-3)+4^(x-5)=68
b/ 1/3-1/3:|2x-1|=-2/3
c/ 2|x-1|-3|x+5|=0
d/( √x +7)^10 =1024*125^2*25^2
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{1024}\right)\)
\(B=4.5^{100}.\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)+1\)