Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có công thức tổng quát của số hạng trong tổng trên có dạng:
\(x_n=\frac{n\left(n+3\right)}{\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n+2-2}{n^2+3n+2}\)
\(=1-\frac{2}{n^2+3n+2}=1-\frac{2}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow\frac{1.4}{2.3}=1-\frac{2}{2.3}\)
\(\frac{2.5}{3.4}=1-\frac{2}{3.4}\)
\(\frac{3.6}{4.5}=1-\frac{2}{4.5}\)
....
\(\frac{98.101}{99.100}=1-\frac{2}{99.100}\)
\(\Rightarrow N=98-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=98-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=98-2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98-1+\frac{1}{50}=97+\frac{1}{50}\)
Vậy 97 < N < 98
ta có A = \(\frac{1.4}{2.3}+\frac{2.5}{3.4}+....+\frac{98.101}{99.100}=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{4950}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{4950}\right)\)(có 98 chữ số 1)
\(=98-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{4950}\right)\)=> A < 98
đi rùi giải tiếp
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
mình vừa mới trả lời xong đấy
Câu hỏi của Do Not Ask Why - Toán lớp 7 - Học toán với OnlineMath
Ta có :
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
A = \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Tách A thành 2 nhóm,ta được :
A = \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Lại có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\text{ }\text{ }\)
\(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\text{ }\text{ }\)
A > \(\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{75}.25+\frac{1}{100}.25\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
A < \(\left(\frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}\right)+\left(\frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}\right)=\frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25\)
\(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(\frac{7}{12}< A< \frac{5}{6}\)