B = 1/52 + 1/62 +......+1/1002
CMR: 1/6 < B < 1/4
Bài 1. Rút gọn
B = 1-7+72-73+...-799
D = 1/3 - 1/32 + 1/33 - 1/34 +...- 1/360
Bài 2. So sánh
A = 1+6+62+...+69 / 1+6+62+...+68
B = 1+5+52+...+59 / 1+5+52+...+58
Cho biểu thức B =1/52 + 1/62 + 1/72 + ... + 1/1002. Chứng tỏ rằng 1/6 < B < 1/4
mn giúp mik với !
Ta có: \(\dfrac{1}{5^2}>\dfrac{1}{5.6};\dfrac{1}{6^2}>\dfrac{1}{6.7};...;\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{96}{505}>\dfrac{1}{6}\) (1)
Ta có: \(\dfrac{1}{5^2}< \dfrac{1}{4.5};\dfrac{1}{6^2}< \dfrac{1}{5.6};\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\) (2)
Từ (1) và (2)⇒\(\dfrac{1}{6}< B< \dfrac{1}{4}\)
B1 a) A=5+53+55+57+..........+5101
b)B=1-1/72-1/73+..........+1/2016
B2 Chứng minh rằng 1/6<1/52+1/62+....+1/1002
B3 Chứng minh rằng 1/n3<1/(n-1)n(n+1)
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Chứng tỏ rằng: B=1/22+1/32+1/42+1/52+1/62+1/72+1/82<1
Đặt B=122+132+...+182B=122+132+...+182A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
=1−18<1(2)=1−18<1(2)
Từ (1);(2)(1);(2) ta có: B<A<1⇒B<1
So sánh A và B. A=1/50+1/51+1/52+...+1/98+1/99
và B= 1/5+1/13+1/14+1/15+1/61?1/62+1/63.
a.Chứng tỏ rằng B = 1/22 + 1/32 + 1/42 + 1/52 + 1/62 + 1/72 +1/82 < 1
b.Cho S = 3/1.4 + 3/4.7 + 3/7.10 +......+3/40.43 + 3/43.46 hãy chứng tỏ rằng S < 1
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!
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)
Tính giá trị biểu thức:
a) A = -11/23.10/-13 +-11/13.3/-23-2/23
b) B = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20
c) C = 52/1.6 + 52/6.11 + 52/11.16 +...+ 52/31.35
d) D = 1. 2 .3 + 2 .4. 6+ 4 .8 .12 + 7 .14 .21/ 1. 3. 5 + 2. 6. 10 + 4 .12 .20 + 7 .21 .35 + 3/5
GIÚP MIK VS AK!
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
Dạng 1: Thực hiện phép tính.
Bài 1: Thực hiện các phép tính rồi phân tích các kết quả ra thừa số nguyên tố.
a, 160 – ( 23 . 52 – 6 . 25 ) b, 4 . 52 – 32 : 24
c, 5871 : [ 928 – ( 247 – 82 . 5 ) d, 777 : 7 +1331 : 113
Bài 2 : Thực hiện phép tính rồi phân tích kết quả ra thừa số nguyên tố:
a, 62 : 4 . 3 + 2 .52 c, 5 . 42 – 18 : 32
Bài 3 : Thực hiện phép tính:
a, 80 - ( 4 . 52 – 3 .23) b, 23 . 75 + 25. 23 + 180
c, 24 . 5 - [ 131 – ( 13 – 4 )2 ] d, 100 : { 250 : [ 450 – ( 4 . 53- 22. 25)]}
So sánh A và B biết rằng A=1/50+1/51+1/52+....+1/98+1/99
B=1/5+1/13+1/14+1/15+1/61+1/62+1/63.
Tính tổng
A = 1 + 5 + 52 + 53 + .... + 52022
B = 6 + 62 + 63 + ..... + 640
C = 42 + 44 + ..... + 414
D = 3 + 33 + 35 + ..... + 325
a: 5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>A=(5^2023-1)/4
b: 6B=6^2+6^3+...+6^41
=>5B=6^41-6
=>B=(6^41-6)/5
c: 16C=4^4+4^6+...+4^16
=>15C=4^16-4^2
=>C=(4^16-4^2)/15
d: 9D=3^3+3^5+...+3^27
=>8D=3^27-3
=>D=(3^27-3)/8