Tính :
\(A=\left(1-\frac{1}{1.2}\right)\left(1-\frac{1}{1.2.3}\right)\left(1-\frac{1}{1.2.3.4}\right)...\left(1-\frac{1}{1.2.3.4.....1986}\right)\)
K
Khách
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Những câu hỏi liên quan
21 tháng 6 2016
bạn làm theo công thức \(\frac{n}{n.\left(n+1\right)}=\frac{n}{n}-\frac{n}{n+1}\)
TM
21 tháng 6 2016
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)
S
18 tháng 7 2018
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3A=\frac{1}{6}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{6\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
=>\(A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+3n^2+3n^2+9n+6-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+6n^2+9n}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
17 tháng 7 2018
Đặt A là biểu thức của đề bài.
Ta có: 3/ 1.2.3.4 = 1/ 1.2.3 -1/ 2.3.4
3/ 2.3.4.5 = 1/ 2.3.4 -1/ 3.4.5
3/ n(n+1)(n+2)(n+3) = 1/ n(n+1)(n+2) -1/ (n+1)(n+2)(n+3)
Do đó: 3A = 1/ 1.2.3 -1/ 2.3.4 + 1/ 2.3.4 - 1/ 3.4.5 +...+ 1/ n(n+1)(n+2) - 1/ (n+1)(n+2)(n+3)
3A = 1/ 1.2.3 - 1/ (n+1)(n+2)(n+3)
3A = 1/6 - 1/ (n+1)(n+2)(n+3)
A = 1/18 - 1/ 3(n+1)(n+2)(n+3)
Đó là kết quả rút gọn. Chúc bạn học tốt.
17 tháng 7 2018
Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
\(\Rightarrow3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}-\frac{1}{\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(A=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
B tự làm nốt nhé
Bài này áp dụng công thức:
\(\frac{a}{b.c.d.e}=\frac{1}{b.c.d}-\frac{1}{c.d.e}\)( đk: \(e-b=a\))
13 tháng 7 2019
\(B=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)\left(n+3\right)}\)
\(B=\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}\right)+\left(\frac{1}{2.3.4}-\frac{1}{3.4.5}\right)+...+\left(\frac{1}{n.\left(n+1\right).\left(n+2\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)\left(n+3\right)}\right)\)
\(B=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}-\frac{1}{\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
\(B=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
11 tháng 8 2017
A= \(\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{19.20.21}-\frac{1}{20.21.22}\right)\)
=\(\frac{1}{3}\left(\frac{1}{6}-\frac{1}{9240}\right)\)
=\(\frac{171}{3080}\)
HC
11 tháng 8 2017
A=1/1-1/2-1/3+1/2-1/3-1/4-1/5+1/3-1/4-1/5-1/6+...+1/19-1/20-1/21-1/22
A=1/1-1/22
A=21/22
Vậy A=21/22
4 tháng 3 2019
Biến đổi ở phân số dạng tổng quát :
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{3}{3n(n+1)(n+2)(n+3)}=\frac{3+n-n}{3n(n+1)(n+2)(n+3)}\)
\(=\frac{1}{3}\left[\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Áp dụng kết quả này vào bài được :
\(\frac{1}{1\cdot2\cdot3\cdot4}=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}\right],\frac{1}{2\cdot3\cdot4\cdot5}=\frac{1}{3}\left[\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}\right],...\)
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Cộng từng vế,ta được : \(S=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
P/S : Xong
4 tháng 3 2019
Ta có: S= \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3S=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(\Rightarrow S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
Vậy \(S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
AH
Akai Haruma
Giáo viên
17 tháng 8
Lời giải:
$3S_n=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{(n+3)-n}{n(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{1.2.3.3}-\frac{1}{3(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}$