Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A=1+A-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}< 1\)
Vậy A <1.
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A=1+A-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}< 1\)
Vậy A <1.
chứng minh rằng :
1/2!.3! + 2/1!.2!.3! + ... + 99/98!.99!.100! < 1
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
chứng minh D=1/2+(1/2)^2+(1/2)^3+....+(1/2)^99<1
a) thu gọn biểu thức sau: a= 5 - 5^2 + 5^3 - 5^4 +...- 5^98 + %^99
b) chứng minh rằng với mọi n thuộc N thì (2^n+1).(2^n+2) đều chia hết cho 3
c) chúng minh: A= 1/1^2 + 1/2^2+ 1/3^2+.....+1/99^2+ 1/100^2 < 1 3/4 (hỗn số)
ChoN=1/2+(1/2)^2+(1/2)^3+(1/2)^4+......+(1/2)^98+(1/2)^99. Chứng minh B<1
chứng minh rằng :
1/ 1! . 2! + 2/ 1! . 2! . 3! + ... + 99/ 98! . 99! . 100! <1
Chứng minh rằng :100- ( 1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
Chứng minh 1/2+(1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^98+(1/2)^99 < 1
A=(1/2+1/2)^2+(1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99
Chứng minh A<1