\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)......\left(\frac{1}{100}-1\right)\)
\(=\left(\frac{1-2}{2}\right).\left(\frac{1-3}{3}\right).\left(\frac{1-4}{4}\right).......\left(\frac{1-100}{100}\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.........\frac{-99}{100}\)
\(=\frac{-\left(1.2.3........99\right)}{2.3.4..........100}=\frac{-1}{100}\)
Chúc bn học tốt
cmr
100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+....+99/100
A=(1+1/2).(1+1/3).(1+1/4)...(1×1/2009)
B=(1-1/2).(1-1/3)...(1-1/100)
B= 1/2.2/3.3/4...99/100
X+1/99+x+2/98+x+3/97+x+4/96
1. Chứng tỏ rằng tổng 100 số đầu tiên của dãy sau nhỏ hơn 1/4:
1/5; 1/45;1/117;1/221;1/357;...
2.tính A/B biết:
A=1/1.300+1/2.301+1/3.302+...+1/101.400
B=1/1.102+1/2.103+...+1/299.400
3.
Chứng minh rằng; 100-(1+1/2+1/3+...+1/100)=1/2+2/3+...+99/100
4. Tính A/B biết : A=1/2+1/3+...+1/200
B=1/199+2/198+...+199/1
5. Tính: 1-1/2+1/3-1/4+...+1/99-1/100 phần 1/51+1/52+...+1/100
tinh
A=1/1+2+1/1+2+3+1/1+2+3+4+...................+1/1+2+3+4+...............+100
1: Tính B
\(B=1+\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{100}\cdot\left(1+2+3+...+100\right)\)
CMR: 1/51 + 1/52 + 1/52 +...+1/100 = 1-1/2 + 1/3 - 1/4 +...+1/99-1/100
cm: 1/2^2+1/3^2+1/4^2+...+1/100^2< 1
Chứng tỏ rằng: 1/2^2+1/3^2+1/4^2+.....+1/100^2<3/4
Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
