1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
1/1.2+1/2.3+1/3.4+...1/100.101
Áp dụng \(\frac{1}{a.\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)
Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
1/1.2+1/2.3+1/3.4+.....+1/100/101
=1-1/2+1/2-1/3+1/3-1/4+.....+1/100-1/101
=1-1/101
=100/101
So sánh A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)với 1
trên violympic phải ko, mình vừa mới giải xong nè
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.... +\frac{1}{100}-\frac{1}{101}\)
triệt tiêu từ từ cuối cùng còn 1 - 1/101 =100/101 = 0,99000000...ĐS: A< 1Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A=1-\frac{1}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A=1-\frac{1}{101}=\frac{100}{101}< 1\)
Vậy : \(A< 1\)
~ Rất vui vì giúp đc bn ~ ^_<
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(A=\left(1-\frac{1}{101}\right)+\left(-\frac{1}{2}+\frac{1}{2}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}< 1\)
=> A <1
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
Tính tổng hoặc hiệu sau:
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+..................+\(\frac{1}{100.101}\)+\(\frac{1}{101.102}\)
B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\)-\(\frac{1}{4.5}\)- .....................-\(\frac{1}{100.101}\)-\(\frac{1}{101.102}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102
B=1-1/102
Tính tổng:
F=\(\frac{1+1.2+3.4+...+100.101}{\left(1.2+2.3+...+99.100\right).2}\)
1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
2.Tìm n
a) A=2n-8/n-1 thuộc z
b) B=3n-5/n-3 thuộc z
c) C=5n-7/2n-3 thuộc z
3. S=12+22+32+...+1002
xin nhờ các bạn giúp đỡ mình chúc các bạn học giỏi
cho A = 1/1.2+1/2.3+1/3.4+...+1/49.50 ; cho B = 1.2+1.3+3.4+....+49.50
tính 50mủ 2A - B/17
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
S=1+2+2^2+2^3+2^4+...+2^100
S=1.2+2.3+3.4+4.5+...+99.100+100.101
Q=1^2+2^2+3^2+...+100^2+101^2
S = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰
2S = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹
S = 2S - S
= (2 + 2² + 2³ + ... + 2¹⁰¹) - (1 + 2 + 2² + ... + 2¹⁰⁰)
= 2¹⁰¹ - 1
------------
S = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 - 1.2.3 + 2
3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102
= 100.101.102
S = 100 . 101 . 102 : 3
= 343400
------------
Q = 1² + 2² + 3² + ... + 100² + 101²
= 101.102.(2.101 + 1) : 6
= 348551
Các bn ơi giúp mk zới
1/1.2+1/2.3+1/3.4+..................+1/100.101
Bn nào trả lời nhanh nhất ,đúng nhất và có lời giải mk tk cho.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+.....\(\frac{1}{100.101}\)
=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+............+\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)