Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt S bằng tổng dãy số trên.
=>S=3/4!+3/5!+.....+3/100!
=>S<3/4!+4/5!+.....+99/100!
=>S<1/3!-1/4!+1/4!-1/5!+.....+1/99!-1/100!
=>S<1/3!-1/100!
=>S<1/3!.Vậy S<1/3!
haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
\(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)
\(D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)
=> \(D< \frac{1}{10}\left(đpcm\right)\)
\(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)
\(D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)
\(= >D< \frac{1}{10}\)
\(\text{k tui}\)
Bạn tham khảo ở link này nhé :
Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
Ta có:
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
Đặt \(I=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
Ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};.....;\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< D\)
Lại có: \(C\cdot D=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\right)\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{10001}\)
Mà C<D \(\Rightarrow C\cdot C< C\cdot D\)
Hay \(C\cdot C< \frac{1}{10001}\)
\(\Rightarrow C< \frac{1}{10001}< \frac{1}{100}\)
Vậy \(C< \frac{1}{100}\left(đpcm\right)\)
Ta có: 1/2^2 < 1/1.2
1/3^2 < 1/2.3
.........................
.......................................
1/100^2 < 1/99.100
Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4
Ta có: 1/2^2 < 1/1.2
1/3^2 < 1/2.3
.........................
.......................................
1/100^2 < 1/99.100
Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4
Đặt :\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
\(N=\frac{2}{3}.\frac{4}{5}...\frac{10000}{10001}\)
Ta thấy:\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};....;\frac{9999}{10000}< \frac{10000}{10001}\)
Mặt khác ta thấy:
\(C.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)
\(C.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{9999}{10000}.\frac{10000}{10001}\)
\(C.N=\frac{1.2.3....9999.10000}{2.3.4....10000.10001}\)
Rút gọn phép tính \(C.N\)
\(C.N=\frac{1}{10001}\)
\(C.C< N\Rightarrow C.C< C.N\)
Hay\(C.C< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{10}.\frac{1}{10}\)
\(\Rightarrow C< \frac{1}{10000}\)(đpcm)
Ta có:
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Mà \(\frac{49}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ta có:\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)(1)
Xét\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)(2)
Mà\(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)(3)
Từ (1), (2), (3)\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Vậy...
Linz
\(D< \frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\)
\(=\frac{1}{1.2.3}-\frac{1}{98.99.100}=\frac{1}{6}-\frac{1}{98.99.10}< \frac{1}{6}\left(ĐPCM\right)\)
T.Anh 2K7(siêu quậy) làm đúng rồi. Làm nhanh và ngắn hơn tớ làm rất rất nhiều!!!