Biết n! = 1.2.3...n (Ví dụ: 3! = 1.2.3 = 6). chứng tỏ rằng S =\(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}< 2\)
Biết n! = 1.2.3. ... . n ( n \(\in\)N* )
Chứng tỏ rằng:
A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)
biết n! = 1.2.3....n (n thuộc N, n\(\ge\)2). chứng tỏ rằng A = \(\frac{1}{2}\)+\(\frac{2}{3}\)+ ....+ \(\frac{2013}{2014}\)< 1
So sánh \(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\) với \(\frac{7}{4}\) (kí hiệu \(n!=1.2.3...n\))
Chứng tỏ:
A = \(\frac{3}{5\cdot2!}\)+ \(\frac{3}{5\cdot3!}\)+\(\frac{3}{5\cdot3!}\)+.....+\(\frac{3}{5\cdot100!}\)< 0.6
(Chú ý : n!=1.2.3.....n(đọc là n giai thừa)
VD: 1!=1 2!=1.2 3!=1.2.3)
chứng tỏ \(\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...100}< 1\)
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
a) 2 +4+6+8+...+2018
= ( 2018+2) x 1009 : 2
= 2020 x 1009 : 2
= 1009 x (2020:2)
= 1009 x 1010
= 1 019 090
b) S = 10 + 102 + 103 + ...+ 10100
=> 10.S = 102 + 103 + 104 +...+ 10101
=> 10.S - S = 10101-10
9.S=10101- 10
\(\Rightarrow S=\frac{10^{101}-10}{9}\)
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5S-S=1-\frac{1}{5^{100}}\)
\(4S=1-\frac{1}{5^{100}}\)
\(S=\frac{1-\frac{1}{5^{100}}}{4}\)
e cx ko nx, e ms hok lp 7 thoy, sang hè ms lp 8! e sr cj nhiều nha!
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
\(S=\frac{1009}{2020}\)
Chứng minh rằng với mọi số nguyên dương n, ta có:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}< \frac{1}{4}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\) \(< \frac{1}{4}\)
Biết n!=1.2.3....n
CMR A=\(\frac{1}{2!}+\frac{2}{3!}+....+\frac{2013}{2014!}< 1\)
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
a;b;c có những câu tương tự rồi, ko giải lại nx
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
b tự làm nốt nha