\(\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2018}\)
Những câu hỏi liên quan
tính \(A=\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.....100}\)
bạn ơi hình như đề bài là:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)thì phải ha.
Đúng 0
Bình luận (0)
chứng tỏ \(\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...100}< 1\)
CM:
1+\(\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)
Đặt A = \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3....n}\)
Ta có: \(\frac{1}{1.2}=\frac{1}{1.2}\)
\(\frac{1}{1.2.3}=\frac{1}{2.3}\)
\(\frac{1}{1.2.3.4}< \frac{1}{3.4}\)
..............
\(\frac{1}{1.2.3....n}< \frac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta được:
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1+1-\frac{1}{n}=2-\frac{1}{n}< 2\)(đpcm)
Đúng 0
Bình luận (0)
\(CM: 1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.4...n}< 2\)
tính \(A=\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.....100}\)
nhanh nhanh mình cần gấp
nếu muốn chứng minh A < 1 thì làm sao
Đúng 0
Bình luận (0)
Tính tổng
\(S=\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+.....+\frac{1}{1.2.3.....50}\)
Mọi người giúp mình vs ạ :D
Tính :
\(A=\left(1-\frac{1}{1.2}\right)\left(1-\frac{1}{1.2.3}\right)\left(1-\frac{1}{1.2.3.4}\right)...\left(1-\frac{1}{1.2.3.4.....1986}\right)\)
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!!!!!!!!!!!!!!!!
bạn làm theo công thức \(\frac{n}{n.\left(n+1\right)}=\frac{n}{n}-\frac{n}{n+1}\)
Đúng 0
Bình luận (0)
a)Đặt A= \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(\Rightarrow2A=1-\frac{1}{2n+1}< 1\)
\(\Rightarrow A< \frac{1}{2}\)(đpcm)
b)Ta có: \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1+1-\frac{1}{n}\)
\(=2-\frac{1}{n}< 2\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}< 2\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)(đpcm)
Đúng 0
Bình luận (0)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}=2A\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=2A\)
\(1-\frac{1}{2n+1}=2A\)
\(\frac{2n+1}{2n+1}-\frac{1}{2n+1}=2A\)
\(\frac{2n}{2n+1}=2A\)
Mà
\(\frac{1}{2}=\frac{1.n}{2.n}=\frac{n}{2n}\)
vì 2A =\(\frac{2n}{2n+1}\)
suy ra A = \(\frac{1n}{2n+1}\)
vì mẫu của \(\frac{1}{2}\)bé hơn mẫu của A
suy ra A < \(\frac{1}{2}\)
suy ra ĐPCM
Đúng 0
Bình luận (0)
CMR : 1+ \(\frac{1}{1.2}\)+\(\frac{1}{1.2.3}\)+\(\frac{1}{1.2.3.4}\)+.....+\(\frac{1}{1.2.3.....n}\)< 2
