Cho:P =(1-1/1+2)(1-1/1+2+3)(1-1/1+2+3+4)...(1-1/1+2+3+...+2014). Tính: 2014/2016P
K
Khách
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14 tháng 4 2015
\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
NH
2
27 tháng 2 2022
\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)
\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)
\(=0\)
DD
Đoàn Đức Hà
Giáo viên
27 tháng 5 2021
\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)
\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)
\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)
\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)
\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)
\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)
\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)
MC
0
biết vậy mà vẫn đòi lấy ảnh! ok!
SN
0
AH
Akai Haruma
Giáo viên
22 tháng 10
Lời giải:
$M=1+\frac{1}{2}.\frac{2(2+1)}{2}+\frac{1}{3}.\frac{3(3+1)}{2}+\frac{1}{4}.\frac{4(4+1)}{2}+....+\frac{1}{2014}.\frac{2014(2014+1)}{2}$
$=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2015}{2}$
$=\frac{2+3+4+....+2015}{2}$
$=\frac{1+2+3+....+2015}{2}-\frac{1}{2}$
$=\frac{2015(2015+1)}{4}-\frac{1}{2}=\frac{2031119}{2}$
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)
\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)
\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)
\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)
\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)