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Ta thấy:
1/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4
........
Coi 1/n = 1/(ax(a+1)) = 1/a - 1/(a+1)
1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/a - 1/(a+1) = 49/50
=1-1/2+1/2-1/3+1/3-......+1/a-1/a+1
Hay A = 1 - 1/(a+1) = 49/50
=> 1/(a+1) = 1 - 49/50
1/(a+1) = 1/50
Vậy (a + 1) = 50 mà n = a x (a+1) => n = (50-1) x 50 = 2450
Ta thấy:
1/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4
........
Coi 1/n = 1/(ax(a+1)) = 1/a - 1/(a+1)
1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/a - 1/(a+1) = 49/50
=1-1/2+1/2-1/3+1/3-......+1/a-1/a+1
Hay A = 1 - 1/(a+1) = 49/50
=> 1/(a+1) = 1 - 49/50
1/(a+1) = 1/50
Vậy (a + 1) = 50 mà n = a x (a+1) => n = (50-1) x 50 = 2450
Bài này phân tích thành :
1/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4
........
1/n = 1/(ax(a+1)) = 1/a - 1/(a+1)
1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/a - 1/(a+1) = 49/50
Hay A = 1 - 1/(a+1) = 49/50
=> 1/(a+1) = 1 - 49/50
1/(a+1) = 1/50
Vậy (a + 1) = 50 mà n = a x (a+1) => n = (50-1) x 50 = 2450
ta co ; 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+..........+1/a-1/b=49/50 ước lượng 1/2; 1/3; 1/3; 1/4; 1/5; 1/6; .........; 1/a = 1-49/50=1/50; vậy n = 50
1/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4 ........ 1/n = 1/(nx(n+1)) = 1/n - 1/(n+1) 1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/n - 1/(n+1) = 49/50 Hay A = 1 - 1/(n+1) = 49/50 => 1/(n+1) = 1 - 49/50 1/(n+1) = 1/50 Suy ra n+1=50 nên n=49
/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4
........
1/n = 1/(nx(n+1)) = 1/n - 1/(n+1)
1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/n - 1/(n+1) = 49/50
Hay A = 1 - 1/(n+1) = 49/50
=> 1/(n+1) = 1 - 49/50
1/(n+1) = 1/50
Suy ra n+1=50 nên n=49
Ta phân tích:
\(\frac{1}{2}\)= \(\frac{1}{1x2}\)= 1 -\(\frac{1}{2}\)
\(\frac{1}{6}\)= \(\frac{1}{2x3}\)= \(\frac{1}{2}\)- \(\frac{1}{3}\)
.....
\(\frac{1}{n}\)= \(\frac{1}{ax\left(a+1\right)}\)= \(\frac{1}{a}\)- \(\frac{1}{a+1}\)
Ta có:A = \(\frac{1}{2}\)+ \(\frac{1}{6}\)+ ... + \(\frac{1}{n}\)= 1 -\(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ ... + \(\frac{1}{a}\)- \(\frac{1}{a+1}\)= \(\frac{49}{50}\)
Hay A = 1 - \(\frac{1}{a+1}\)= \(\frac{49}{50}\)
\(\Rightarrow\) \(\frac{1}{a+1}\)= 1 -\(\frac{49}{50}\)
\(\Rightarrow\)\(\frac{1}{a+1}\)= \(\frac{1}{50}\)
Vậy (a + 1) = 50 mà n = a x (a+1) => n = (50-1) x 50 = 2450
Ta lấy \(\frac{49}{50}\)trừ đi 5 phân số kia
Sau đó sẽ là phân số .........
Vậy là tìm được n
ta có dạng tổng quát sau : 1/ 2 = 1/(2*1)
1/6 = 1/(2*3)
1/12 = 1/(3*4)
....................
1/n = 1/(x-1)x
cộng vế theo vế ta có :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....+\frac{1}{x\left(x-1\right)}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{\left(x-1\right)}-\frac{1}{x}\)
\(=1-\frac{1}{x}\)
Mà A = 49/50
Nên \(1-\frac{1}{x}=\frac{49}{50}\)
\(\frac{1}{x}=1-\frac{49}{50}=\frac{1}{50}\)
\(x=50\)
\(n=x\left(x-1\right)=50\times49=2450\)
Vậy n = 2450
<=> A = 1/1.2 + 1/2.3 + 1/3.4 + ..... + 1/a.( a + 1 ) = 49/50 [ a.( a + 1 ) = n ]
<=> A = 1/1 - 1/2 + 1/2 - 1/3 + ....... + 1/a - 1/a + 1 = 49/50
<=> A = 1 - 1/a + 1 = 49/50
<=> A = 1/a + 1 = 1 - 49/50
<=> A = 1/a + 1 = 1/50
=> a + 1 = 50
=> n = 50. ( 50 - 1 ) = 2450
Vậy n = 2450
1/2 = 1/(1x2) = 1 - 1/2
1/6 = 1/(2x3) = 1/2 - 1/3
1/12 = 1/(3x4) = 1/3 - 1/4
........
1/n = 1/(ax(a+1)) = 1/a - 1/(a+1)
1 /2 + 1/6 + 1/12 + 1/20 + 1/30 +...+ 1/n = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+1/a - 1/(a+1) = 49/50
Hay A = 1 - 1/(a+1) = 49/50
=> 1/(a+1) = 1 - 49/50
1/(a+1) = 1/50
Vậy (a + 1) = 50 mà n = a x (a+1) => n = (50-1) x 50 = 2450