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\(\frac{1}{3}\times\frac{3}{5}\times\frac{5}{7}\times...\times\frac{99}{101}\times\frac{101}{103}\)
\(=\frac{1\times3\times5\times...\times99\times101}{3\times5\times7\times...\times101\times103}\)
\(=\frac{1}{103}\)
\(B\text{=}\dfrac{3}{1\times3}+\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+...+\dfrac{3}{99\times101}\)
\(B\text{=}3\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(\dfrac{3-1}{1\times3}+\dfrac{5-3}{3\times5}+...+\dfrac{101-99}{99\times101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(1-\dfrac{1}{101}\right)\)
\(B\text{=}\dfrac{300}{202}\)
Đặt S = | 1 | + | 1 | + … + | 1 |
1 . 3 | 3 . 5 | 99 . 101 |
1 | - | 1 | = | 3 - 1 | = | 2 |
1 | 3 | 1 . 3 | 1 . 3 |
1 | = | 1 | ( | 1 | - | 1 | ) |
1 . 3 | 2 | 1 | 3 |
1 | = | 1 | ( | 1 | - | 1 | ) |
3 . 5 | 2 | 3 | 5 |
1 | = | 1 | ( | 1 | - | 1 | ) |
5 . 7 | 2 | 5 | 7 |
1 | = | 1 | ( | 1 | - | 1 | ) |
99 . 101 | 2 | 99 | 101 |
S = | 1 | ( | 1 | - | 1 | ) |
2 | 1 | 101 |
S = | 1 | 101 - 1 | |
2 | 101 |
S = | 100 |
202 |
Tổng ban đầu = | 50 |
101 |
\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
\(D=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{99}{101}\)
\(D=\frac{1\cdot3\cdot5\cdot7\cdot...\cdot99}{3\cdot5\cdot7\cdot9\cdot...\cdot101}\)
\(D=\frac{1}{101}\)
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A = \(\frac{1}{1\cdot3}\)+ \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ ..... + \(\frac{1}{99.101}\)
= \(\frac{1}{2}\). ( \(\frac{1}{1.3}\)+ \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ ...... + \(\frac{1}{99.101}\))
= \(\frac{1}{2}\). ( 1 - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ........ + \(\frac{1}{99}\)- \(\frac{1}{101}\))
= \(\frac{1}{2}\). ( 1 - \(\frac{1}{101}\))
= \(\frac{1}{2}\). \(\frac{100}{101}\)= \(\frac{50}{101}\)
Thấy đúng thì cho mình một k nha!!!
Khoảng cách giữa 2 số hạng liên tiếp:
3-1=2
Số lượng số hạng:
(101-1):2 + 1= 51 (số hạng)
Tổng trên bằng:
(101+1):2 x 51=2601
Đ.số: 2601
SSH= \(\dfrac{101-1}{2}+1=51\)
Tổng: \(\dfrac{\left(101+1\right)\cdot51}{2}=2601\)