Tính : D =\(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+...+2016}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
8 tháng 9 2016
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{2015}{2017}\right)}\)
\(\Rightarrow A=2017\)
8 tháng 9 2016
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{2}{2016.2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+\frac{2015}{2017}}\)
\(A=2017\)
LB
30 tháng 9 2016
Ta có:1+2+3+...+n=\(\frac{n.\left(n+1\right)}{2}\)
=>B=\(\frac{2.2016}{\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)
=>B=\(\frac{2016}{\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}}\)
=>B=\(\frac{2016}{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{2016}-\frac{1}{2017}\right)}\)
=>B=
7 tháng 9 2016
Mẫu số của A \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(2016+1\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
MA
7 tháng 9 2016
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+..+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{2016.2017}\right)}\) .
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}=\frac{4032}{1+\frac{2015}{2017}}\)
\(A=2017\)
7 tháng 9 2016
Mẫu số \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(0+1\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
AH
Tính
- \(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+2016}}\)
1
19 tháng 10 2016
Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\)
\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{1}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=\frac{2.2016}{2017}\)
Vậy phân số đề bài cho \(=\frac{2.2016}{\frac{2.2016}{2017}}=2.2016.\frac{2017}{2.2016}=2017\)
ND
2
LM
13 tháng 11 2016
mẫu:\(\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\)
=\(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\)
=\(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)
=\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2016.2017}\right)\)
=\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
=\(1+2\left(\frac{1}{2}-\frac{1}{2017}\right)\)
=\(1+1-\frac{2}{2017}\)
=\(\frac{4032}{2017}\)
=>Biểu thức:\(\frac{4032}{\frac{4032}{2017}}\)
=\(2017\)
14 tháng 11 2016
Ta có công thức tổng quát với n tự nhiên là
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)
Áp dụng công thức vào bài toán ta được
\(\frac{2.2016}{\frac{1}{1}+\frac{1}{1+2}+..+\frac{1}{1+2+...+2016}}=\frac{2.2016}{\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{2016.2017}}\)
\(=\frac{2.2016}{2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\right)}=\frac{2.2016}{2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(=\frac{2.2016}{2\left(1-\frac{1}{2017}\right)}=\frac{2.2016}{\frac{2.2016}{2017}}=2017\)
6 tháng 9 2016
Xét mẫu số của F :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2016\cdot2017}{2}}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{2017}\right)=2-\frac{2}{2017}=\frac{4032}{2017}\)
Suy ra : \(F=\frac{2.2016}{\frac{4032}{2017}}=\frac{2.2016.2017}{4032}=2017\)