Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
K
Khách
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28 tháng 1 2016
=\(\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+........+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{7-3}{7.3}+\frac{11-7}{7.11}+........+\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{7}{7.3}-\frac{3}{7.3}+\frac{11}{7.11}-\frac{7}{7.11}+......+\frac{4n+3}{\left(4n-1\right)\left(4n+3\right)}-\frac{4n-1}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7.}-\frac{1}{11}+......+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3-3}{3\left(4n+3\right)}\right)\)
\(=\frac{5}{4}.\frac{4n}{3\left(4n+3\right)}=\frac{4.n.5}{3\left(4n+3\right).4}=\frac{5n}{3\left(4n+3\right)}\)
ban nen xem lai dau bai di minh giai dung 100% do
ma neu dau bai ra nhu ket qua cua to thi tick cho minh nha
27 tháng 1 2017
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
30 tháng 1 2017
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
30 tháng 1 2017
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
J
2 tháng 7 2018
\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)
Mk thực sự nghĩ đề hình như bị sai hay sao ấy!
MH
22 tháng 8 2016
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)
\(\Rightarrow10x=3.\left(3x+3\right)\)
\(\Rightarrow10x=9x+9\)
\(\Rightarrow x=9\)
Vậy...
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