Chứng minh bất đẳng thức sau:
a/ \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right).\left(2n+1\right)}<\frac{1}{2}\)
b/ \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}<2\)
Giúp mình với!!!!!!!!!!!!!!!!
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
Tính
E=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{ }{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
Tính
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{47.48.49.50}\)
Tính nhanh :
a ) S = 2+4+6+8+.....+2018
b ) S= 10+102+103+...+10100
c) S=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.....+\frac{1}{5^{100}}\)
d) S=\(\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+....+\frac{2018!}{2020!}\)
biết rằng : n!=1×2×3×...×n
VD : 1! = 1
2! = 2.1
3!=1.2.3
4!=1.2.3.4
! : giai thừa
Tính nhanh :
a ) S = 2+4+6+8+.....+2018
b ) S= 10+102+103+...+10100
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+....+\frac{2018!}{2020!}\)
biết rằng : \(n!=1\times2\times3\times...\times n\)
VD : 1! = 1
2! = 2.1
3!=1.2.3
4!=1.2.3.4
! : giai thừa
tính
a) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+.....+\frac{1}{37.38.39.40}\)
b)\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{16}{3^{16}}\)
TÍNH TỔNG:
\(S=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
