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
NN
0
NM
3
9 tháng 5 2017
chịu mẹ kiếp toán 7 cho vào đề kiểm tra toán 6 ai mà lm dc
9 tháng 5 2017
=1-1/4+1-1/9+1-1/16+...+1-1/10000
=(1+1+1+...+1)+(-1/4-1/9-1/16-...-1/10000)
=99+(-1/4-1/9-1/16-...-1/10000)
Vì 99+(-1/4-1/9-1/16-...-1/10000)>98
=>C>98
Vây C>98
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 11 2018
Đặt \(A=\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}=1-\dfrac{1}{4}+1-\dfrac{1}{9}+...+1-\dfrac{1}{10000}\)
\(=99-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)=99-B\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>0\Rightarrow99-B< 99\Rightarrow A< 99\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}\)
\(\Rightarrow A=99-B>99-\left(1-\dfrac{1}{100}\right)=98+\dfrac{1}{100}>98\)
Vậy \(98< \dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}< 99\)
NT
0
LM
0
LM
1
AH
Akai Haruma
Giáo viên
16 tháng 7 2023
Lời giải:
$A=(1-\frac{1}{4})+(1-\frac{1}{9})+(1-\frac{1}{16})+....+(1-\frac{1}{10000})$
$=(1+1+...+1)-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})$
$=99-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})< 99$
NT
0
VN
20 tháng 4 2023
Tham khảo :
3.98.1615.....100009999
=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}=2.21.3.3.32.4.4.43.5.....100.10099.101
=\dfrac{\left(1.2.3.....99\right)}{\left(2.3.4.....100\right)}.\dfrac{\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)}=(2.3.4.....100)(1.2.3.....99).(2.3.4.....100)(3.4.5.....101)
=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}=1001.2101=200101
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 ~