1.
a, chứng tỏ
1/2^2+1/3^2+...+1/2017^2<1
b,1/4+1/16+1/36+1/64+1/100+1/144+...+1/10000<1/2
c,cho A=1/2^2+1/3^2...+1/9^2
chứng tỏ:2/5<a<8/9
d,chứng tỏ:A=1+1/2^2+...+1/100^2<1/3/4
e,chứng tỏ:1/2^2+1/3^2+...+1/100^2<1
A=1/21+1/22+1/23+....+1/40
Chứng tỏ1/2<A<1
A=1/2+1/22+1/23+...1/2100 chứng tỏ0<A<1
A=1/21+1/22+1/23+1/24+...1/40 chứng tỏ1/2<A<1
khó wá
Ta có:A=1/21+1/22+1/23+...+1/40(có 20 số hạng)
A>1/40+1/40+...+1/40
A>20/40=1/2(1)
A=1/21+1/22+1/23+...+1/40(có 20 số hạng)
A<1/20+1/20+1/20+...+1/20
A<20/20=1(2)
Từ (1) và (2)=>1/2<A<1
Ta có :A=1/2+1/2^2+1/2^3+...+1/2^100
2A=1+1/2+1/2^2+...+1/2^99
2A-A=(1+1/2+1/2^2+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)
A=1-1/2^100
Dễ thấy A>0 và 1-1/2^100<1
=>0<A<1
CHo A=1/2^2 +1/3^2+...1/2018^2 . Chứng minh 2017/4038 >A>2017/2018
Chữa đề \(\frac{2017}{4038}< A< \frac{2017}{2018}\)
Ta có: \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\Leftrightarrow A< 1-\frac{1}{2018}=\frac{2017}{2018}\)(1)
Lại có: \(A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{2019}=\frac{2017}{4038}\)(2)
Từ (1) và (2) => đpcm
Chứng minh A là một số nguyên dương :
A = \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+\)\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+...+\left(\frac{1}{2017}\right)^2\)
Cho F = 3/10+3/11+3/12+3/13+3/14
Chứng tỏ1<F<2
Có F = \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
\(\Rightarrow F=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{15}{14}>1\left(1\right)\)
\(F=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< 2=\frac{20}{10}\left(2\right)\)
Từ ( 1 ),( 2 ) \(\Rightarrow1< F< 2\)
Vậy \(1< F< 2\left(đpcm\right)\)
P.S: đpcm: Điều phải chứng minh
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
F < \(\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}=2\)
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
cho A= 1/2^2 + 1/3^2 + ... + 1/2017^2. Chứng minh A<3/4
Ta có : \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
.....................
\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=\frac{1}{4}+\frac{1}{2}-\frac{1}{2017}\)
\(A=\frac{3}{4}-\frac{1}{2017}\left(đpcm\right)\) . Vậy A < \(\frac{3}{4}\)
chứng tỏ A=1/2^2+1/3^2+1/4^2+...+1/2017^2+1/2018^2<1
chứng tỏ A=1/2^2+1/3^2+1/4^2+...+1/2017^2+1/2018^2<1