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Ta có:
\(M=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...\)
Dựa vào quy luật trên=>Số hạng thứ 30 là:\(\frac{1}{98.101}\)
\(\Rightarrow3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{98}-\frac{1}{101}\)
\(\Rightarrow M=\left(1-\frac{1}{101}\right):3=\frac{100}{101}.\frac{1}{3}=\frac{100}{303}\)
Mình viết hơi tắt mong bạn thông cảm.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(52:x=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}\)
\(52:x=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}\)
\(52:x=\frac{1}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}\right)\)
\(52:x=\frac{1}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(52:x=\frac{1}{3}\times\left(1-\frac{1}{13}\right)\)
\(52:x=\frac{1}{3}\times\frac{12}{13}\)
\(52:x=\frac{4}{13}\)
\(x=52:\frac{4}{13}\)
\(x=169\)
52 :x =1/4 + 1/28 +1/70 + 1/130
52 : x = 4/13
x = 52 : 4/13
x = 169
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)
\(=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}+\frac{1}{13\times16}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
Chúc bạn học tốt ^^
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(\frac{1}{4}=\frac{1}{1\times4}\) ; \(\frac{1}{28}=\frac{1}{4\times7}\) ; \(\frac{1}{70}=\frac{1}{7\times10}\) ; \(\frac{1}{130}=\frac{1}{10\times13}\) ; ...
Từ đó ta suy ra được quy luật của dãy là nghịch đảo tích của 2 số liên tiếp trong dãy sau:
1 ; 4 ; 7 ; 10 ; 13 ; ...
Số thứ 100 trong dãy trên là:
\(1+3\times99=298\)
Vậy phân số thứ 100 trong dãy trên là:
\(\frac{1}{298\times\left(298+3\right)}=\frac{1}{298\times301}=\frac{1}{89698}\)
Vậy phân số thứ 100 là \(\frac{1}{89698}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{304}.\)
\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{16.19}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{13-10}{10.13}+.....+\frac{19-16}{16.19}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{16}-\frac{1}{19}\)
\(=1-\frac{1}{19}=\frac{18}{19}\)
3/4+3/28+3/70+3/130+...+3/304
= 3 /1.4 + 3/4.7 + 3/7.10 + 3/10.13 +....+ 3 /16.19
= 1 -1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ...+1/16 - 1/19
= 1 - 1/19
=18/19