Các câu hỏi tương tự
chứng minh
1/22+1/32+1/42+1/52+...+1/1002 >3/4
M = 1002– 992 + 982 – 972 + … + 22 – 12;
N = (202+ 182 + 162 + … + 42 + 22) – (192 + 172 + 152 + … + 32 + 12);
P = (-1)n.(-1)2n+1.(-1)n+1.
B1 a) A=5+53+55+57+..........+5101
b)B=1-1/72-1/73+..........+1/2016
B2 Chứng minh rằng 1/6<1/52+1/62+....+1/1002
B3 Chứng minh rằng 1/n3<1/(n-1)n(n+1)
1/22 + 1/42 + 1/62 + 1/82 + ... + 1/20222 < 1/2 (gần gấp)
cmr: 1 - 1/2 + 1/3 -1/4 +...-1/200 + 1/2001 - 1/2002 = 1/1002 +...+ 1/2002
AI GIÚP NHÉ!!! THANKS:)
let S be 1!(12+1+1)+2!(22+2+1)+3!(32+3+1)+...+100!(1002+100+1). Find S+1/101!.(as usual, k! = 1.2.3.....(k-1).k)
a, chứng minh rằng : 1-1/2+1/3-1/4+...-1/2000+1/2001-1/2002 = 1/1002+ ...+ 1/2002
giúp mk nha
Biết
1)\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\).CM: \(a^2=b\cdot c\)
2)CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{\text{4}}+...+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
chứng minh : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+.....+\frac{1}{2002}\)