A=(1-1/4).(1-1/9).(1-1/16).(1-1/25)...(1-1/10000)
So sánh A= 1/4+1/9+1/16+1/25+.....+1/10000 và 3/4
A=1/(2x2)+1/(3x3)+...+1/(100x100)
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1)
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
so sánh A= 1/4+1/9+1/16+1/25+.....+1/10000 và 3/4
A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
cho A=1/4+1/9+1/16+1/25+...+1/10000 so sanh A voi 3/4
So sánh K và 1
K=1/4+1/9+1/16+1/25+...+1/10000
chung to rang ;1/4+1/9+1/16+1/25+...+1/10000 <1
S=1/4+1/9+1/16+1/25+...+1/10000 ngày mai mình nộp bài rồi
So sánh : 1/4+1/9+1/16+1/25+...+1/10000 và 3/4
Cách giải nhé.
A = 1+4+9+16+25+...+10000 = ?
So sanh 1/4+1/9+1/16+1/25+1/36+...+1/10000 voi 1 va neu cach giai(so sanh) nua nhe!
\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+.....+\frac{1}{100.100}\)
\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}<\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}=\frac{99}{100}<1\)
Vậy \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}<1\)