Cho S = \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\). Chứng tỏ rằng 1/6 < S < 1/4
Chứng minh rằng:
\(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ta có:
\(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)\(=\frac{1}{4}-\frac{1}{100}=\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)
Ta có : \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)
\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)
\(\frac{1}{7^2}=\frac{1}{7.7}< \frac{1}{6.7}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow S< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(S< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(S< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}=\frac{24}{100}\)
Mà \(\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow S< \frac{1}{2}\)
Vậy S<\(\frac{1}{2}\).
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2017}}\)
chứng tỏ A<1
2,
\(S=2+2^2+2^3+...+2^{99}\)
C/t: S chia hết cho 7, 31
3,
\(A=1+5+5^2+5^3+5^4+5^5+...+5^{99}+5^{100}\)
Tính A
4,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)<1
5,
CHỨNG tỏ rằng các p/s tối giản vs mọi số tự nhiên n
a,\(\frac{n+1}{2n+3}\)b,\(\frac{2n+3}{4n+8}\)
6,
a,TÍnh A và B
A=\(\frac{2016^{2016}+2}{2016^{1016}-1}\)B=\(\frac{2016^{2016}}{2016^{2016}-3}\)
b, tính
C=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
LÀm NHANH Hộ MK ,MAi mk Phải Nộp.
2/
S = 2 + 22 + 23 +...+ 299
= (2+22+23) +...+ (297+298+299)
= 2(1+2+22)+...+297(1+2+22)
= 2.7 +...+ 297.7
= 7(2+...+297) chia hết cho 7
S = 2+22+23+...+299
= (2+22+23+24+25)+...+(295+296+297+298+299)
= 2(1+2+22+23+24)+...+295(1+2+22+23+24)
= 2.31+...+295.31
= 31(2+...+295) chia hết cho 31
3/
A = 1+5+52+....+5100 (1)
5A = 5+52+53+...+5101 (2)
Lấy (2) - (1) ta được
4A = 5101 - 1
A = \(\frac{5^{101}-1}{4}\)
4/
Đặt A là tên của biểu thức trên
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
........
\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy...
5/
a, Gọi UCLN(n+1,2n+3) = d
Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d
2n+3 chia hết cho d
=> 2n+2 - (2n+3) chia hết cho d
=> -1 chia hết cho d => d = {-1;1}
Vậy...
b, Gọi UCLN(2n+3,4n+8) = d
Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d
4n+8 chia hết cho d
=> 4n+6 - (4n+8) chia hết cho d
=> -2 chia hết cho d => d = {1;-1;2;-2}
Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}
Vậy...
6/
a,Vì B > 1
\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A < B
b, C = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
bài 1,Cho \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\).Hãy chứng tỏ rằng B>1.
bài 2,Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\).Hãy chứng tỏ rằng S<1.
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
chứng minh rằng\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.......+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng :
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng :
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
chứng minh rằng:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+.......+\frac{1}{100^2}< \frac{1}{4}\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)
Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)
Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)
Từ (1) và (2) =>đpcm
Chứng minh rằng : \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ae nhớ viết lời giải chi tiết nhé. Ko mình đập cho k trượt phát lào đó !
Ta có:
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\)
\(-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
=> đpcm
Thank bn Hoàng đạo thứ 7 nhé. Cho 3 k r nhé hihi
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{8^2}<\frac{1}{7.8}\)
<=> B<\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\)
<=> B<\(\frac{1}{1}-\frac{1}{2}+.......+\frac{1}{7}-\frac{1}{8}\)
<=> B<\(1-\frac{1}{8}\)
<=> B<\(\frac{7}{8}\) <1