Tính tổng : A=38/25 + 9/10 - 11/15 + 13/21 - 15/28 + 17/36 - ................+197/4851-199/4950
Tính tổng : A=38/25 + 9/10 - 11/15 + 13/21 - 15/28 + 17/36 - ................+197/4851-199/4950
bỏ tạm 38/25 ra
=(9/10-11/15) +(13/21-15/28) +(197/4851-199/4950)
=1/6+1/12+1/20+...+1/49*50=2450
=3-2/2x3+4-3/3x4+5-4/4x5+....50-49/49x50
=1/2-1/3+1/3-...-1/50
1/2-1/50=12/25
=38/25+12/25=50/25
=2
Ta nhóm các số hạng từ thứ hai trở đi thành từng cặp: ( C = \frac{38}{25} + \left( \frac{9}{10} - \frac{11}{15} \right) + \left( \frac{13}{21} - \frac{15}{28} \right) + ... + \left( \frac{197}{4851} - \frac{199}{4950} \right) ) Tính tổng mỗi cặp: Cặp tổng quát thứ (n): ( \frac{2n+7}{(n+1)(2n+3)} - \frac{2n+9}{(n+2)(2n+3)} ) cho (n=1, 2, 3, ... ) Với (n=1): ( \frac{9}{10} - \frac{11}{15} = \frac{1}{6} = \frac{1}{2 \times 3} ) Với (n=2): ( \frac{13}{21} - \frac{15}{28} = \frac{1}{12} = \frac{1}{3 \times 4} ) Với (n=48): ( \frac{197}{4851} - \frac{199}{4950} = \frac{1}{2450} = \frac{1}{49 \times 50} ) Tổng các cặp là ( S' = \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ... + \frac{1}{49 \times 50} = \sum_{n=2}^{49} \frac{1}{n(n+1)} = \sum_{n=2}^{49} \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{2} - \frac{1}{50} = \frac{24}{50} = \frac{12}{25} ). ( C = \frac{38}{25} + \frac{12}{25} = \frac{50}{25} = 2 ). Đáp án: 2
Tính tổng \(A=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
bỏ tạm 38/25 ra
=(9/10-11/15) +(13/21-15/28) +(197/4851-199/4950)
=1/6+1/12+1/20+...+1/49*50=2450
=3-2/2x3+4-3/3x4+5-4/4x5+....50-49/49x50
=1/2-1/3+1/3-...-1/50
1/2-1/50=12/25
=38/25+12/25=50/25
=2
(x-2)^2=38/25+9/10-11/15+13/21-15/28+17/36-...+197/4851-199/4950
Tính tổng:
\(S=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
bỏ tạm 38/25 ra
=(9/10-11/15) +(13/21-15/28) +(197/4851-199/4950)
=1/6+1/12+1/20+...+1/49*50=2450
=3-2/2x3+4-3/3x4+5-4/4x5+....50-49/49x50
=1/2-1/3+1/3-...-1/50
1/2-1/50=12/25
=38/25+12/25=50/25
=2
38/25+9/10-11/15+13/21-15/28+17/36_............+197/4851-199/4950
(C=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950})
bỏ tạm 38/25 ra
=(9/10-11/15) +(13/21-15/28) +(197/4851-199/4950)
=1/6+1/12+1/20+...+1/49*50=2450
=3-2/2x3+4-3/3x4+5-4/4x5+....50-49/49x50
=1/2-1/3+1/3-...-1/50
1/2-1/50=12/25
=38/25+12/25=50/25
=2
38/25+9/10-11/15+13/21-15/28+17/36_............+197/4851-199/4950
Tìm x, biết :
(x - 2)^2 = 38/25 + 9/10 - 11/15 + 13/21 - 15/28 + 17/36 - ..... + 197/4851 - 199/4950
Tìm x:
[x-2]2=38/25+9/10-11/15+13/21-15/28+17/36-...+197/4851-199/4950
Tính tổng : A=38/25 + 9/10 - 11/15 + 13/21 - 15/28 + 17/36 - ................+197/4851-199/4950
A= 38/25+9/10-11/15+13/21-15/28+17/36-...+197/4851-199/4950
Ta có thể viết lại : (9/20-11/15)+(13/21-15/28)+17/36-19/45...+(197/4851-199/4950)
Ta thấy:(9/10-11/15)=1/6=1/2x3=1/2-1/3
(13/21-15/28)=1/12=1/3x4=1/3-1/4
(17/36-19/45)=1/20=1/4x5=1/4-1/5
............................
Ta được:9/10-11/15+13/21-15/28+17/36-19/45...+197/4851-199/4950=1/2-1/3+1/3-1/4+1/4-1/5=24/50
Vậy:A= 38/25+24/50
A=2