cm 3/4+8/9+15/16+......+9999/10000<99
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 4 2017
Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+................+\dfrac{9999}{10000}\)
\(A=\dfrac{1.3}{2^2}+\dfrac{2.4}{3^2}+\dfrac{3.5}{4^2}+....................+\dfrac{99.101}{100^2}\)
\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+..................+\dfrac{100^2-1}{100^2}\)
\(A=\dfrac{2^2}{2^2}-\dfrac{1}{2^2}+\dfrac{3^3}{3^2}-\dfrac{1}{3^2}+............+\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\)
\(A=\left(\dfrac{2^2}{2^2}+\dfrac{3^3}{3^3}+...........+\dfrac{100^2}{100^2}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^3}+........+\dfrac{1}{100^2}\right)\)
\(A=\left(1+1+........+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^3}+............+\dfrac{1}{100^2}\right)\)
\(A=99-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..........+\dfrac{1}{100^2}\right)\)
Ta có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+............+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)\(\dfrac{1}{2^2}+........+\dfrac{1}{100^2}< \dfrac{1}{1}-\dfrac{1}{2}+.......+\dfrac{1}{99}-\dfrac{1}{100}\)\(\Rightarrow\dfrac{1}{2^2}+.........+\dfrac{1}{100^2}< 1-\dfrac{1}{100}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{100^2}< \dfrac{100}{101}\)
\(\Rightarrow99-\left(\dfrac{1}{2^2}+...........+\dfrac{1}{100^2}\right)< 99-\dfrac{100}{101}\)
\(\Rightarrow A< 99-\dfrac{100}{101}\)
\(\Rightarrow a< 99\rightarrowđpcm\)
~ Học tốt ~
NA
Nguyễn An Ninh
VIP
1 tháng 5 2023
3/4.8/9.15/16...9999/10000
=\(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\)...\(\dfrac{99.101}{100.100}\)
=\(\dfrac{1.2...99}{2.3.100}\).\(\dfrac{3.4...101}{2.3.100}\)
=\(\dfrac{1}{100}\).\(\dfrac{101}{2}\)
=\(\dfrac{101}{200}\)
2 tháng 3 2022
\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}\)+...+\(\frac{9999}{10000}\)
= (1-\(\frac{1}{4}\)) +(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{10000}\))
= 99 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)) => 99 - A
Dễ thấy A>0 =>S < 99 (1)
Lại có A= \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)
=> A<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...\(\frac{1}{99}\)-\(\frac{1}{100}\)
=>A<1-\(\frac{1}{100}\)<1
...
26 tháng 2 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(=\frac{1.101}{100.2}=\frac{101}{200}\)
AT
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2021
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
DQ
21 tháng 3 2017
= 3 . 8 . 15 .... 9999 / 4 . 9 . 16 .... 10000
= ( 1 . 3 ) . ( 2 . 4 ) .( 3 . 5) .... ( 99 .... 101 ) / ( 2. 2) . (3.3). (4.4)...(100.100)
= 1. 101/100.2
= 101/ 200
k nha , đúng đó
21 tháng 3 2017
1*3/2*2.2*4/3*3.3*5/4*4.....99*101/100*100. =1*2*3*...*99/2*3*4*...*100.3*4*5*...*101/2*3*4*...*100. =1/100 . 101/2. =101/200.
7 tháng 4 2019
3/4 . 8/9 . 15/16 ... 9999/10000
= 1.3/2.2 . 2.4/3.3 ... 99.101/100.100
= 1 . 2 . ... . 99 / 2 . 3 . 100 × 3 . 4 ... 101 / 2 . 3 ... 100
= 1 / 100 . 101 / 2
= 101 / 200
7 tháng 4 2019
=1.3/2.2 .2.4/3.3 .3.5/4.4 . ...... 99.101/100.100
=1.2.3.4.5 ...... .99/2.3.4.....100 . 3.4.5 ....... .101/2.3.4.5 .... .100
=1/100 .101/2
=101/200
k cho mink nha
TB
0