Chứng minh rằng:
\(\frac{3}{4!}+\frac{3}{5!}+\frac{3}{6!}+...+\frac{3}{100!}<\frac{1}{3!}\)
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Những câu hỏi liên quan
6 tháng 2 2020
\(D< \frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\)
\(=\frac{1}{1.2.3}-\frac{1}{98.99.100}=\frac{1}{6}-\frac{1}{98.99.10}< \frac{1}{6}\left(ĐPCM\right)\)
WH
2 tháng 5 2018
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 \(C< \frac{1}{100}\left(đpcm\right)\)
2 tháng 5 2016
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
SH
2 tháng 5 2016
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
9 tháng 6 2017
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
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
TL
1
22 tháng 6 2017
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)
DP
19 tháng 7 2017
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
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!
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