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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/3*5+1/5*7+1/7*9+...+1/97*99
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=1/3-1/99
32/99
![](https://rs.olm.vn/images/avt/0.png?1311)
\(G=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+.....+\frac{1}{95.97.99}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+....+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}.\frac{3200}{9603}=\frac{800}{9603}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
số số hạng là
(99-1):2 + 1= 50(số)
tổng là
(99+1) x 50 : 2= 2500
đap số 2500
số số hạng :
(99-1):2+1=50(số hạng)
tổng:
(99+1)*50:2=2500
k cho chi nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{97.99}\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{3}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{3}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{32}{99}\)
\(=\frac{16}{33}\)
b)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chép nhầm đề rồi
Đề đúng là
99-97+95-93+91-89+... +7-5+3-1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,1 + 1,2 + 1,3 +... + 1,7 + 1,8 + 1,9
= (1,1 + 1,9) + (1,2 + 1,8) + (1,3 + 1,7) + (1,4 + 1,6) + 1,5
= 3 + 3 + 3 + 3 + 1,5
= 13,5
1 + 2 + 3 + 4 + ..... + 97 + 98 + 99 + 100
= (1 + 100) + (2 + 99) + (3 + 98) + .... + (54 + 57) + (55 + 56)
= 101 + 101 + 101 + ..... + 101 (50 số 101)
= 101 x 50 = 5050
1 + 3 + 5 + 7 + 9 + ...... + 93 + 95 + 97 + 99
= (1 + 99) + (3 + 97) + (5 + 95) + ..... + (49 + 51)
= 100 + 100 + 100 + .... + 100 (25 số 100)
= 100 x 25 = 2500
lấy 1.1+1.9+1.2+1.8+1.3+1.7+1.4+1.6+1.5=5.5
cái kia cũng làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{4}{3x5}+\dfrac{4}{5x7}+\dfrac{4}{7x9}+...+\dfrac{4}{97x99}+\dfrac{4}{99x101}\)
\(A=4x\left(\dfrac{1}{3x5}+\dfrac{1}{5x7}+\dfrac{1}{7x9}+...+\dfrac{1}{97x99}+\dfrac{1}{99x101}\right)\)
\(A=4x\left[\dfrac{1}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}x\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\dfrac{1}{2}x\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\dfrac{1}{2}x\left(\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}x\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right]\)
\(A=4x\dfrac{1}{2}x\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right]\)
\(A=2x\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2x\dfrac{98}{303}=\dfrac{916}{303}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+3+5+7+9 +.........+97+99
=\(\frac{99.\left(99+1\right)}{2}\)
=9900/2
=4950
Khoảng cách giữa các số là:
3- 1= 2
Số các số hạng là:
( 99- 1)/ 2+ 1= 50
Tổng các số là:
( 99+ 1)* 50/ 2= 2500
Vậy A= 2500.
Tổng A gồm có:
( 99-1):(3-1)+1=50(số hạng)
Tổng A là:
( 99+1)x50:2=2500