Tính
a ) A = 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 149.150 b ) B = 5 1.3 + 5 3.5 + 5 5.7 + ... + 5 201.203
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{149.150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{149}-\dfrac{1}{150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{150}\)
\(A=\dfrac{150}{150}-\dfrac{1}{150}\)
\(A=\dfrac{149}{150}\)
Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
=>\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=>\(A=1-\frac{1}{50}\)
=>\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)
A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)
A=1- 1 + \(\dfrac{1}{98}\)
A= \(\dfrac{1}{98}\)
Lời giải:
$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$
$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$
$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$
$=1-\frac{1}{98}$
$\Rightarrow A=\frac{1}{98}$
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
Lời giải :
Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101
3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3
=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)
=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102
=100.101.102
S=100.101.34=343400
1.Tính
a) Ta có:
A=(1-1/22).(1-1/32)...(1-1/1002)
=>A=3/22.8/32.....9999/1002
=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)
=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)
=>A=1/100.101/2
=>A=101/200
b) Ta có:
B=-1/1.2-1/2.3-1/3.4-...-1/100.101
=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)
=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=>B=-(1-1/101)
=>B=-100/101
c) Ta có:
C=1.2+2.3+3.4+...+100.101
=>3C=1.2.3+2.3.3+3.4.3+...+100.101.3
=>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)
=>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102
=>3C=100.101.102
=>3C=1030200
=>C=343400
Chúc bạn hok tốt nhé >:)!!!!!
=> Ta thấy rằng mỗi số hạng trong dãu số trên đều là tích của hai số tự nhiên liên tiếp, khi đó:
Gọi a1 = 1.2 → 3a1 = 1.2.3 → 3a1 = 1.2.3 - 0.1.2
Tương tự:
a2 = 2.3 → 3a2 = 2.3.3 → 3a2 = 2.3.4 - 1.2.3
a3 = 3.4 → 3a3 = 3.3.4 → 3a3 = 3.4.5 - 2.3.4 ....
a(n - 1) = (n - 1).n → 3a(n - 1) = 3(n - 1)n → 3a(n - 1) = (n - 1).n.(n + 1) - (n - 2).(n - 1).n
an = n.(n - 1) → 3an = 3n(n + 1) → 3an = n(n + 1)(n + 2) - (n - 1)n(n + 1)
Cộng vế với vế của các đẳng thức trên ta được:
3(a1 + a2 + a3 +...+ an) = n(n + 1)(n + 2)
-> A = n.(n+1) .( n+2) / 3
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+....+n\left(n+1\right).3\)
\(\Leftrightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(\Leftrightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(\Leftrightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
A= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)]
=n(n+1)(n+2)
=>A
ta có : A = 1.2 + 2.3 + 3.4 + ...... + n(n + 1)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + n(n + 1)(n + 2)
=> 3A = n(n + 1)(n + 2)
=> A = n(n + 1)(n + 2)/3
a) A = 149 150 b) B = 505 203