cmr:\(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+....+\dfrac{2011}{1.2.3....2012}< 1\)
cmr:\(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+\dfrac{3}{1.2.3.4}+....+\dfrac{2011}{1.2...2012}< 1\)
Chứng minh BĐT sau
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}< \dfrac{1}{2}\)
b)
a)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)
P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!
cmr:1/1.2+2/1.2.3+3/1.2.3.4+....+2011/1.2...2012<1
Chứng minh bất đẳng thức sau:
\(1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+...+\dfrac{1}{1.2.3.....n}< 2\)
Cho S = \(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+\dfrac{3}{1.2.3.4}+....+\dfrac{99}{1.2.3.....99.100}\)
Chứng minh rằng : S<1
\(S=\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+........+\dfrac{99}{1.2.......100}\)
\(=\dfrac{1}{2!}+\dfrac{2}{3!}+....+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+.......+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+....+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
A=\(\left(1-\dfrac{1}{1.2}\right)\left(1-\dfrac{1}{1.2.3}\right).........\left(1-\dfrac{1}{1.2.3....n}\right)\)
1) Tìm 2 số nguyên tố x, y sao cho: \(x^2-6y^2=1\)
2) Cho \(B=1.2.3...2012.\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)
CMR: B chia hết cho 2013
1) \(x^2-6y^2=1\)
=> \(x^2-1=6y^2\)
=> \(y^2=\frac{x^2-1}{6}\)
Nhận thấy y^2 thuộc Ư của \(\dfrac{x^2-1}{6}\)
=> \(y^2\) là số chẵn.
Mà y là số nguyên tố.
=> y = 2.
Thay vào:
=> \(x^2-1=\dfrac{4}{6}=24\)
=> \(x^2=25\)
=> \(x=5\)
Vậy: x = 5; y = 2.
Chứng tỏ: \(\dfrac{1}{1.2}\) + \(\dfrac{1}{1.2.3}\) +\(\dfrac{1}{1.2.3.4}\)+.......+\(\dfrac{1}{1.2.3........100}\) <1
1.Tính
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
B=\(\dfrac{3}{5.6}+\dfrac{3}{6.7}+\dfrac{3}{7.8}+.....+\dfrac{3}{101.102}\)
C=\(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}\)
D=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)