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Khoảng cách giữa hai thừa số trong mỗi số hạng là 2, nhân 2 vế của A với 3 lần khoảng cách này ta được :
6A=1.3.6 + 3.5.6 + 5.7.6 + ... + 97.99.6
=1.3(5+1) + 3.5(7-1) + 5.7(9-3) + ... + 97.99(101-95)
=1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99
=1.3.5 + 3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7+ ... + 97.99.101 - 97.97.99
=3+97.99.101
\(\frac{1+97.33.101}{1}=161651\)
Ta có :
B = 1.3 + 3.5 + 5.7 + 7.9 + ... + 97.99
6.B = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6
6.B = 1.3.[ 5 - (-1) ] + 3.5.( 7 - 1 ) + 5.7.( 9 - 3 ) + ...+ 97.99.( 101 - 95 )
6.B = 1.3.5 - ( -1).3.5 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99
6.B = 97.99.101 - ( -1 ) .3.5
6.B = 97.99.101 + 1.3.5
6.B = 969918
=> B = 161653.
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1.3 + 3.5 |+ 5.7 + ... + 97.99
6A = 1.3.6 + 3.5.(7-1) + 5.7.(9-3) + ... + 97.99.(101-95)
6A = 1.3.6 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99
6A = 1.3.6 + 97.99.101 - 1.3.5
6A = 3.(1 + 97.33.101)
2A = 1 + 323301 = 323302
A = 161651
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=1.3+3.5+5.7+.....+95.97+97.99\)
\(\frac{2}{B}=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{95.97}+\frac{2}{97.99}\)
\(\frac{2}{B}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{97}-\frac{1}{99}\)
\(\frac{2}{B}=\frac{1}{1}-\frac{1}{99}=\frac{90}{99}=\frac{30}{33}\)
\(B=\frac{2}{\frac{30}{33}}=\frac{2.33}{30}=\frac{33}{15}\)
\(6A=1.3.6+3.5.6+5.7.6+...+97.99.6\)
= \(1.3\left(5+1\right)+3.5\left(7-1\right)+5.7\left(9-3\right)+...97.99\left(101-95\right)\)
= \(.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+97.99.101-97.97.99\)
= 3 + 97 .99 . 101
= \(\frac{1+97.33.101}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{99}\)
\(B=\dfrac{99}{99}-\dfrac{1}{99}\)
\(B=\dfrac{98}{99}\)
#YVA
B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)
B=\(\dfrac{98}{99}:2\)
B=\(\dfrac{49}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tham khảo nhé!
Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101
A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)
A = (12 + 32 + 52 + … + 972 + 992) + 2.(1 + 3 + 5 + … + 97 + 99).
Đặt B = 12 + 32 + 52 + … + 992
=> B = (12 + 22 + 32 + 42 + … + 1002) – 22.(12 + 22 + 32 + 42 + … + 502)
Tính dãy tổng quát C = 12 + 22 + 32 + … + n2
C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]
C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)
C = = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6
Áp dụng vào B ta được:
B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650
=> A = 166650 + 2.(1 + 99).50 : 2
=> A = 166650 + 5000 = 172650.
Đ/s: A = 172650.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Kết quả là:
161651
Đpá số: 161651
161651
ban nhe
tk nha@@@@@@@@@@@@@@@@@@@@@@@@2
hihi