1/3+3/3.7+5/3.7.11+...+2n+1/3.7.11.....(4n+3)<1/2
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Ta có: S = \(\dfrac{1}{3}+\dfrac{3}{3.7}+\dfrac{5}{3.7.11}+...+\dfrac{2n+1}{3.7.11...\left(4n+3\right)}\)
⇒ 2S = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+2}{3.7.11...\left(4n+3\right)}\)
⇒ 2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+3}{3.7.11...\left(4n+3\right)}\)
Đến đây nó sẽ rút gọn liên tục và sau nhiều lần rút gọn ta có:
2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+\dfrac{1}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{11}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{1}{3.7}\) = \(\dfrac{2}{3}+\dfrac{7}{3.7}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
Suy ra 2S < 1 ⇒ S < \(\dfrac{1}{2}\)(đpcm)
8C=8/3.7.11+8/7.11.15+8/11.15.19+8/15.19.23+8/19.23.27+8/23.27.31
8C=1/3.7-1/7.11+1/7.11-1/11.15+1/11.15-1/15.19+1/15.19-1/19.23+1/19.23+-1/23.27-1/27.31
8C=1/3.7-1/27.31
8C=1/21-1/837
8C=272/5859
C=34/5859
Bài 1:
a: =-3/7
b: =-3/5
c: =1/13
d: =-1/2
Bài 2:
a: \(=\dfrac{21\cdot11}{22\cdot9}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)
b: \(=\dfrac{49\cdot8}{10}=\dfrac{196}{5}\)
\(A=\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}\)
\(\frac{4}{5}.A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right).\left(4n+3\right)}\)
\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)
\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{4n+3}\)
\(\frac{4}{5}.A=\frac{4n+3}{12n+9}-\frac{3}{12n+9}\)
\(\frac{4}{5}.A=\frac{4n}{12n+9}\)
\(A=\frac{4n}{12n+9}:\frac{4}{5}\)
\(A=\frac{4n}{12n+9}.\frac{5}{4}\)
\(A=\frac{5n}{12n+9}\)
Đề bài sai nha bn
Ủng hộ mk nha ^_^
\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)
Mk thực sự nghĩ đề hình như bị sai hay sao ấy!
nếu lm sai thì thôi na
đặt vế trái lak A
4A=5(5/3.7+5/7.11+.....+5/(4n-1)(4n+3)
4A=(1/3-1/4n+3)
4A=4n+3-3/3(4n+3)
4A=4/12n+9
A=5/4n+3