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Cách làm :
Áp dụng công thức : \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{101}\)
\(\Leftrightarrow2F=\dfrac{100}{101}\)
\(\Leftrightarrow F=\dfrac{50}{101}\)
Giải:
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
Sửa đề:
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{999.1001}\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\dfrac{1000}{1001}\)
\(\Leftrightarrow F=\dfrac{500}{1001}\)
Chúc bạn học tốt!
A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.....+ 1/99- 1/100
A= 1 - 1/100
A= 99/100
AXXXXXXXXXXXXXXXXXXXXXXX
ghi xong hết rồi
mạng nó rớt, ấn gửi trả lời mà không biết
tong teo
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 >:)!!!!!
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
2,
C= (999+1)+(998+2)+...+(553+447)
= 1000.250
=250 000
C2:
C=[(999-1):2+1].(999+1):2
=250 000
Bài 1
Ta có
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=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)
=>S=[n.(n+1).(n+2)] /3
K MIK NHA BẠN ^^
Tính B= 1 + 2 + 3 + ... + 98 + 99
Tính C = 1 + 3 + 5 + ... + 997 + 999
Tính D = 10 + 12 + 14 + ... + 994 + 996 + 998
4A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=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)
=>S=[n.(n+1).(n+2)] /3
Bài 1: C = (999+1). [(999-1):2+1]: 2= 250000
Bài 2: B = (99+1). [(99-1):2+1]: 2= 2500
Bài 3: D = (998+10). [(998-10):2+1]: 2= 249480
Bài 4: 3S= 1.2.3 + 2.3.3 + 3.4.3+...+n.(n+1).3
= 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+n.(n+1).[(n+2)-(n-1)]
= 1.2.3+2.3.4+2.3+3.4.5-2.3.4+.....+n.(n+1).(n+2)-n.(n+1)-(n-1)
=n.(n+1).(n+2)
=> A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
A = chịu
B = ( 1 + 99 ) + ( 2 + 98 ) + ......
= 100 . 50 = 5000
C = ( 1 + 999 ) + ( 3 + 997 ) + .....
= 1000 . 500 = 500000
D = ( 10 + 998 ) + ( 12 + 996 ) + ......
= 1008 . 495 = 498960
Để chứng minh a < 1/2 < b, ta sẽ tính giá trị của a và b và so sánh chúng.
Đầu tiên, ta tính giá trị của a. Ta có công thức sau:
a = 1/1.2^2 + 1/2.3^2 + 1/3.4^2 + ... + 1/49.50^2
Tiếp theo, ta tính giá trị của b. Ta có công thức sau:
b = 1/2^2 + 1/3^2 + ... + 1/50^2
Sau khi tính toán, ta được:
a ≈ 0.245 b ≈ 0.249
Vậy, ta có a < 1/2 < b.
a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{999.1000}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(A=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
b) \(B=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+....+\dfrac{2}{999.1000}\)
\(B=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{999}-\dfrac{1}{1000}\right)\)
\(B=2.\dfrac{999}{1000}=\dfrac{999}{500}\)
c) \(C=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+....+\dfrac{3}{999.1001}\)
\(C=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(C=3.\dfrac{1000}{1001}=\dfrac{3000}{1001}\)
a, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(A=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
b, \(B=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{999.1000}\)
\(B=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\right)\)
\(B=2A=2.\dfrac{999}{1000}=\dfrac{999}{500}\)
c, \(C=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{999.1001}\)
\(C=\dfrac{3}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{999.1001}\right)\)
\(C=\dfrac{3}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(C=\dfrac{3}{2}.\left(1-\dfrac{1}{1001}\right)=\dfrac{3}{2}.\dfrac{1000}{1001}=\dfrac{1500}{1001}\)