Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{9^2}\)
Chứng tỏ \(\frac{2}{5}< S< \frac{8}{9}\)
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\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}\)
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}<\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(S<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(S<\frac{1}{2}-\frac{1}{20}<\frac{1}{2}\)
Vậy \(S<\frac{1}{2}\)
1/2+1/4+1/8+1/16+1/32+1/64=32/64+16/64+8/64+4/64+2/32+1/64=63/64<1
Xét \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}a>0\)
Ta có: \(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
Vì a>0, D>0 nên \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng ta có: \(D=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)=100-\frac{1}{100}=99,99\)
\(\frac{x-12}{3}=\frac{x+1}{4}\)
=>(x-12).4=(x+1)*3
4x-48=3x+3
4x-3x=48+3
x=51
(x-12)/3=(x+1)/4
(x-12)*4=(x+1)*3
x*4-12*4=x*3+1*3
4x-48=3x+3
4x-3x=3+48
x=51
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
\(\frac{1}{3^2}<\frac{1}{3.4}\)
\(\frac{1}{4^2}<\frac{1}{4.5}\)
\(\frac{1}{5^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)
Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)
hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2
Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100
A<1/2-1/100<1/2
Ta có điều phải chứng minh.
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
M=1/10 + 1/15 + 1/21 +....+ 1/120
M=2/20 +2/30+2/42+....+2/240
M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16
M=2.(1/4.5 +......+ 1/15.16)
M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)
M=2.(1/4 - 1/16)
M=2.(4/16 - 1/16)
M=2. 3/16
M=6/16=3/8
Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M
Có 1/2 = 4/8>3/8 =>1/2 >M
=> 1/3 < M < 1/2
1/2^2>1/2.3;1/3^2>1/3.4;......;1/9^2>1/9.10
suy ra S > 1/2.3+1/3.4+......+1/9.10
S> 1/2-1/3+1/3-1/4 +.....+1/9-1/10
S> 1/2-1/10=2/5
Vay 2/5 < S
Vậy còn S < \(\frac{8}{9}\)thì sao, bạn quên chưa chứng minh rồi