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27 tháng 4 2021
a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)
\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)
Vậy ta có biểu thức:
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)
Vậy B < 1 (đpcm)
27 tháng 4 2021
Giải:
a) Ta có:
1/22=1/2.2 < 1/1.2
1/32=1/3.3 < 1/2.3
1/42=1/4.4 < 1/3.4
1/52=1/5.5 < 1/4.5
1/62=1/6.6 < 1/5.6
1/72=1/7.7 < 1/6.7
1/82=1/8.8 <1/7.8
⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
B<1/1-1/8
B<7/8
mà 7/8<1
⇒B<7/8<1
⇒B<1
b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46
S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=45/46
Vì 45/46<1 nên S<1
Vậy S<1
Chúc bạn học tốt!
Q
1
YN
8 tháng 4 2022
`Answer:`
\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)
a) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)
b) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)
\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)
\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)
\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)
\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)
LH
2
16 tháng 4 2022
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5 Chúc bạn học tốt !
KD
2
YN
10 tháng 2 2023
\(S=\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+...+\dfrac{1}{103^2}\)
\(\Rightarrow2S=\dfrac{2}{5^2}+\dfrac{2}{7^2}+\dfrac{2}{9^2}+...+\dfrac{2}{103^2}\)
Có:
\(\dfrac{2}{5^2}=\dfrac{2}{5.5}< \dfrac{2}{4.6}=\dfrac{1}{4}-\dfrac{1}{6}\)
\(\dfrac{2}{7^2}=\dfrac{2}{7.7}< \dfrac{2}{6.8}=\dfrac{1}{6}-\dfrac{1}{8}\)
\(\dfrac{2}{9^2}=\dfrac{2}{9.9}< \dfrac{2}{8.10}=\dfrac{1}{8}-\dfrac{1}{10}\)
...
\(\dfrac{2}{103^2}=\dfrac{2}{103.103}< \dfrac{1}{102.104}=\dfrac{1}{102}-\dfrac{1}{104}\)
\(\Rightarrow2S< \dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\)
\(\Rightarrow2S< \dfrac{25}{104}\)
\(\Rightarrow S< \dfrac{25}{208}< \dfrac{5}{32}\)
\(\Rightarrow S< \dfrac{5}{32}\).
S
10 tháng 2 2023
Ta có:
\(\dfrac{1}{5^2}< \dfrac{1}{4.6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6.8}\)
\(\dfrac{1}{9^2}< \dfrac{1}{8.10}\)
\(...\)
\(\dfrac{1}{103^2}< \dfrac{1}{102.104}\)
\(\Rightarrow S\)\(< \dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)\(\left(1\right)\)
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{102.104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}.\dfrac{25}{104}\)
\(=\dfrac{25}{208}< \dfrac{25}{160}\)\(\left(2\right)\)
Mà \(\dfrac{25}{160}=\dfrac{5}{32}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow S< \dfrac{5}{32}\)
AU
2
20 tháng 4 2022
S=(1/31+1/32+1/33+...+1/40)+(1/41+1/42+1/43+...+1/50)+(1/51+1/52+1/53+...+1/60)"10 sống hạng mỗi ngoặc"
S<1/30 x 10+1/40 x 10+1/50 x 10
S<1/3+1/4+1/5=47/60<48/60=4/5
Học tốt~
NH
0
AN
0
NM
0
NH
1