1 + 1+ 2 + 2 + 3 =
Chứng minh rằng:
a,A=1/2+1/2^2+1/2^3+.+1/2^2<1
b,B=1/3+1/3^2+1/3^3+...+1/3^n<1/2
c,B=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2015-1/2^2016<1/3
d,D=1/3+2/3^2+3/3^3+4/3^4+...+100/3^100<3/4
?reeeeeeeeeeee
Ủa, cái số gì đây??????
Tính:
2 - 1 = 3 - 1 = 1 + 1 = 1 + 2 =
3- 1 = 3 - 2 = 2 - 1 = 3 - 2 =
3 -2 = 2 - 1 = 3 - 1 = 3 - 1 =
2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
ok nhá
Tính:
1 + 2 = … | 3 – 1 = … | 1 + 1 = … | 2 – 1 = … |
3 – 2 = … | 3 – 2 = … | 2 – 1 = … | 3 – 1 = … |
3 – 1 = … | 2 – 1 = … | 3 – 1 = … | 3 – 2 = … |
Lời giải chi tiết:
1 + 2 = 3 | 3 – 1 = 2 | 1 + 1 = 2 | 2 – 1 = 1 |
3 – 2 = 1 | 3 – 2 = 1 | 2 – 1 = 1 | 3 – 1 = 2 |
3 – 1 = 2 | 2 – 1 = 1 | 3 – 1 = 2 | 3 – 2 = 1 |
1+2=3 | 3-1=2 | 1+1=2 | 2-1=1 |
3-2=1 | 3-2=1 | 2-1=1 | 3-1=2 |
3-1=2 | 2-1=1 | 3-1=2 | 3-2=1 |
#HT#
1. 3, 2, 2, 1.
2. 1 ,1 ,1 ,2.
3. 2, 1, 2, 1.
HT!
Tính:
A=(1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4)...(1-1/1+2+3+4+...+2022)
B=1+1/2(1+2)+1/3(1+2+3)+1/100(1+2+3+...+100)
Tính nhanh
1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+3+2+1+2+3+1+2+3=
Đọc kĩ đề
1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+2+3+1+2+3
= 6+6+6+6+6+6+6+6+6
= 6x9
= 54
>, <, = ?
2 + 2 … 5 | 2 + 1 … 1 + 2 | 3 + 1 … 3 + 2 |
2 + 3 … 5 | 2 + 2 …1 + 2 | 3 + 1 … 1 + 3 |
5 + 0 … 5 | 2 + 0 …1 + 2 | 1 + 4 …. 4 + 1 |
Lời giải chi tiết:
2 + 2 < 5 | 2 + 1 = 1 + 2 | 3 + 1 < 3 + 2 |
2 + 3 = 5 | 2 + 2 > 1 + 2 | 3 + 1 = 1 + 3 |
5 + 0 = 5 | 2 + 0 < 1 + 2 | 1 + 4 = 4 + 1 |
2 + 2 < 5 2 + 1 = 1 + 2 3 + 1 < 3 + 2
2 + 3 = 5 2 + 2 > 1 + 2 3 + 1 = 1 + 3
5 + 0 = 5 2 + 0 < 1 + 2 1 + 4 = 4 + 1
+, - ?
1… 2 = 3 | 2 … 1 = 3 | 1…1 = 2 | 1…4 = 5 |
3…1 = 2 | 3…2 = 1 | 2…1 = 1 | 2…2 = 4 |
Lời giải chi tiết:
1 + 2 = 3 | 2 + 1 = 3 | 1 + 1 = 2 | 1 + 4 = 5 |
3 - 1 = 2 | 3 - 2 = 1 | 2 - 1 = 1 | 2 + 2 = 4 |
Chứng minh rằng:
a) A=1/2^2+1/3^2+1/4^2+...+1/2010^2<1
b) B=1/2+2/2^2+3/2^3+...+100/2^100<2
c) C=1/3+2/3^2+3/3^3+...+100/3^100<3/4
d) D=1/2^3+1/3^3+1/4^3+...+1/n^3<1/4 (n€ N;n> hoặc = 3)
e) E=1/3^3+1/4^3+1/5^3+...+1/n^3<1/12 (n€N; n> hoặc = 3)
f) F=2/1*4/3*6/5*...*200/199<20
g) G=3/4+5/36+7/144+...+2n+1/n^2*(n+1)^2<1 (n nguyên dương)
h) H=1/2*(1/6+1/24+1/60+...+1/9240)>57/462
i) I=1/31+1/32+1/33+...+1/2048>3
j) J=(1-1/3)*(1-1/6)*(1-1/10)*...*(1-1/253)<2/5
k) K=1/2!+2/3!+3/4!+...+n-1/n! (n€N;n> hoặc = 2)
l) L=1/2!+5/3!+11/4!+...+n^2+n-1/(n+1)!<2
m) 1/6M=1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
Có thể mình hơi phũ tí nhưng mình bảo đảm một thế kỉ sau sẽ không ai ngồi giải hết đống bài này cho bạn đâu, hỏi từng câu thôi
P/s: chắc bạn đánh mỏi tay lắm
Ta có: D<1/1.2.3+1/2.3.4+1/3.4.5+...+1/(n-1).n.(n+1)
D<1/2.(2/1.2.3+2/2.3.4+2/3.4.5+...+2/(n-1).n.(n+1))
D<1/2.(1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/(n-1).n-1/n.(n+1))
D<1/2.((1/2-1/n.(n+1))
D<1/4-1/2.n.(n+1)<1/4
D<1/4
Tính?
1 + 1 =… | 1 + 2 =… | 2 + 2 = … | 1 + 1 =… |
2 + 1 =… | 1 + 3 =… | 3 + 1 =… | 1 + 2 =… |
3 + 1 =… | 1 + 1 =… | 1 + 3 =… | 2 + 1 =… |
Lời giải chi tiết:
1 + 1 = 2 | 1 + 2 = 3 | 2 + 2 = 4 | 1 + 1 = 2 |
2 + 1 = 3 | 1 + 3 = 4 | 3 + 1 = 4 | 1 + 2 = 3 |
3 + 1 = | 1 + 1 = 2 | 1 + 3 = 4 | 2 + 1 = 3 |
Tính tổng sau: a) 1/2+1/6+1/12+1/20+1/30 b) 1/15+1/35+1/63+1/99+1/143 c) 1/6+1/12+1/20+1/30+1/42+1/56 d) 1/2+1/2^2+1/2^3+1/2^4+1/2^5 e) 1/7+1/7^2+1/7^3+...+1/7^100 f) 1+1/2*(1+2)+1/3*(1+2+3)+1/4*(1+2+3+4)+...+1/200*(1+2+3+..+200) g) (1/2+1)*(1/3+1)*(1/4+1)*..*(1/100+1) h) (1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/2022) Giúp mk vs ạkkk
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)