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2/3+2/15+2/35+...+2/n x (n+2)=322/323
=2/1.3+2.3.5+2/5.7+...+2/n x(n+2)=322/323
= n+2 =323
n=323-2
n=321
2/3+2/15+2/35+...+2/n x (n+2) = 322/323
2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323
1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323
1-1/n+2 = 322/323
1/n+2 = 1-322/323
1/n+2 = 1/323
=> n+2 = 323
n = 323 - 2 = 321
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+2}{x+2}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+1}{x+2}=\frac{332}{323}\)
=>332.(x+2)=323.(x+1)
=>332x+664=323x+323
=>332x-323x=323-664
=>x.(332-323)=-323
=>9x=-323
=>x=-323/9
vậy n=-323/9 .(-323/9+2)=98515/81
Nếu 2/3 + 2/15 + 2/35 + ................+2 /A = 60/61.Tìm A
Giải
Đặt B = 2/3 + 2/15 + 2/35 + ................+2 /A
= 2/(1x3) + 2/3x5 + 2/5x7 + ... + 2/a x (a+2) (trong đó A = a x (a +2)
= 3/1x3 - 1/1x3 + 5/3x5 - 3/3x5 +7/5x7 - 5/5x7 +... + (a+2)/a x (a + 2) - a/a x (a+2)
=1 - 1/3+1/3 -1/5+1/5 - ... - 1/(a + 2)
= 1 - 1/(a + 2)
Vậy 1 - 1/(a + 2) = 60/61
=> (a + 1)/ (a + 2) = 60/61
=> a = 59
Mà A = a x (a +2) nên A = 59 x (59 + 2) = 59 x 61 = 3599
ĐS: A = 3599
=> 2/1x3 +2/3x5+2/5x7+2/7x9+...+2/nx(n+2)
=>1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9+...+1/n-1/n+2
=>1-1/n+2=100/101
1/n+2=1-100/101
1/n+2=1/101
=>n+2=101
=>n=101-2
=>n=99
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