Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow1< 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
Mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2\Rightarrow A< 2\left(đpcm\right)\)
b, B = 2 + 22 + 23 +...+ 230
= (2+22+23+24+25+26)+...+(225+226+227+228+229+230)
= 2(1+2+22+23+24+25)+...+225(1+2+22+23+24+25)
= 2.63+...+225.63
= 63(2+...+225)
Vì 63 chia hết cho 21 nên 63(2+...+225) chia hết cho 21
Vậy B chia hết cho 21
b, Dãy B có 30 số ta nhóm thành 5 nhóm mỗi nhóm gồm 6 số thì mỗi nhóm đều chia hết cho 63
Ví dụ : 2^1+2^2+2^3+2^4+2^5+2^6
= 2.(1+2+2^2+2^3+2^4+2^5)
= 2.63 chia hết cho 6
=> B chia hết cho 63
Mà 63 = 21.3 nên B chia hết cho 21
a, Có : A = 1 + 1/2.2 + 1/3.3 + ....... + 1/50.50
=> A < 1 + 1/1.2 + 1/2.3 + ...... + 1/49.50
= 1 + 1 - 1/2 + 1/2 - 1/3 + ...... + 1/49 - 1/50
= 1 + 1 - 1/50
= 2 - 1/50
< 2
=> A < 2
Tk mk nha
b, 21 + 22 + 23 + ... + 230
= ( 21 + 22 + 23 + 24 + 25 + 26 ) + ( 27 + 28 + 29 + 210 + 211 + 212 ) + ... + ( 225 + 226 + 227 + 228 + 229 + 230 )
= 21 . ( 20 + 21 + 22 + 23 + 24 + 25 ) + 27 . ( 20 + 21 + 22 + 23 + 24 + 25 ) + ... + 225 . ( 20 + 21 + 22 + 23 + 24 + 25 )
= 2 . 63 + 27 . 63 + ... + 225 . 63
= 63 . ( 2 + 27 + ... + 225 )
= 21 . 3 . ( 2 + 27 + ... + 225 ) \(⋮\)21
a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A< 1+\left(1-\frac{1}{100}\right)\Rightarrow A< 1+1-\frac{1}{100}\Rightarrow A< 2-\frac{1}{100}\Rightarrow A< 2\left(ĐPCM\right)\)
b, \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2011\cdot2012}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Rightarrow B< 1-\frac{1}{2012}\Rightarrow B< 1\left(1\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{2013}\Rightarrow\frac{1}{2}-\frac{1}{2013}< B\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{2}-\frac{1}{2013}< B< 1\)
a)A=1+1/22+1/32+....+1/1002
<1+1/1.2+1/2.3+...+1/99.100=1+1-1/2+1/2-1/3+...+1/99-1/100=2-1/100=199/200<2
b)B=1/22+1/32+...+1/20122
<1/1.2+1/2.3+...+1/2011.2012=1-1/2+1/2-1/3+...+1/2011-1/2012=1-1/2012=2011/2012
1/2-1/2013=2011/4026<2011/2012<1
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
a)=>A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt tổng trong ngoặc là M
=>M=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)\(=1-\frac{1}{50}< 1\)
Khi đó A=1+M (M<1)
Ta có công thức :1+x<2 nếu x<1
=>A<1
bn ơi A < 2 makk