biết mỗi câu a

ukm thế cx dc

cau a bang 101/200, k cho minh nha'

A=1.2.3+2.3.4+....+99.100.101

4A=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+....+98.99.100.(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-3.4.5.2+....+98.99.100.101-98.99.100.97

4A=98.99.100.101

4A=97990200

A=97990200/4

A=24497550

B=1.2+3.4+5.6+7.8+8.9+...+999.1000

3B=1.2.3+2.3.(4-1)+3.4(5-2)+....+998.999(1001-998)

3B=1.2.3+2.3.4-2.3.1+3.4.5-3.4.2+....+998.999.1001-998.999.998

3B=999.1000.1001

3B=999999000

B=999999000/3

B=333333000

C=1+4+9+16+25+36+.....+10000

C=1^2+2^2+3^2+4^2+5^2+6^2+....+100^2

C=(1^2+3^2+5^2+.....+99^2)+(2^2+4^2+6^2+....+100^2)

C=99.100.101/6  +   100.101.102/6

C=166650         +171700

C=338350

Còn câu d bạn  dựa vào câu c là làm được ngay bây h mk mỏi tay rùi ko muốn đánh nữa khi nào rảnh mk gửi công thức cho nha bây h mk bận rùi.

chúc bn học tốt

   A=1.2.3+2.3.4+....+99.100.101

4.A=1.2.3.(4-0)+2.3.4.(5-1)+...+99.100.101.(102-98)

4.A=1.2.3.1-0.1.2.3+2.3.4.5-1.2.3.4+....+99.100.101.102-98.99.100.101

4.A=99.100.101.102

  A=\(\frac{99.100.101.102}{4}\)

   B=1.2+2.3+3.4+...+999.1000

3.B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+999.1000.(1001-998)

3.B=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+999.1000.1001-998.999.1000

3.B=999.1000.1001

=>B=\(\frac{999.1000.1001}{3}\)

C và D dễ lắm bạn tự làm nhé

mình lại có cách làm khác là tìm số thừa ra lấy số cuối cộng số đầu chia cho  2 . tính số các số hạng là số cuối trừ số đầu chia khoảng cách cộng 1  , roi tinh gia tri 1 cap la so cuoi cong so dau sau do tinh so cap la lay so cac so hang chia 2 va di tinh tong thi ban tu lam

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)

\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)

\(B=\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot...\cdot\left(1-\dfrac{1}{1326}\right)\\ =\dfrac{20}{21}\cdot\dfrac{27}{28}\cdot\dfrac{35}{36}\cdot...\cdot\dfrac{1325}{1326}\\=\dfrac{40}{42}\cdot\dfrac{54}{56}\cdot\dfrac{70}{72}\cdot...\cdot\dfrac{2650}{2652}\\ =\dfrac{5\cdot8}{6\cdot7}\cdot\dfrac{6\cdot9}{7\cdot8}\cdot\dfrac{7\cdot10}{8\cdot9}\cdot...\cdot\dfrac{50\cdot53}{51\cdot52}\\ =\dfrac{5\cdot8\cdot6\cdot9\cdot7\cdot10\cdot...\cdot50\cdot53}{6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot...\cdot51\cdot52}\\ =\dfrac{\left(5\cdot6\cdot7\cdot...\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot...\cdot53\right)}{\left(6\cdot7\cdot8\cdot...\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot...\cdot52\right)}=\dfrac{5\cdot53}{51\cdot7}=\dfrac{265}{357} \)

a) A = 2 + 4 + 6 + 8 + ... + 1000

Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )

               = (1000 + 2) . 500 : 2 = 250500

c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

Bạn giỏi bạn làm đi đã ngu zồi thích tỏ ra minh ngu hơn. Bạn sợ bạn nếu ko nói câu đấy người ta tưởng bạn khôn chắc

\(\)\(A=\frac{3}{1\times3}+\frac{3}{3\times5}+...+\frac{3}{49\times51}\)

\(\Leftrightarrow A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(\Leftrightarrow A=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)

\(\Leftrightarrow A=\frac{3}{2}.\frac{50}{51}\)

\(\Leftrightarrow A=\frac{25}{17}\)

\(\)\(\)

Chi tiết cho mk nha

\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)

\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)

\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)

\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)

\(=\frac{100\cdot2}{1\cdot101}\)

\(=\frac{200}{101}\)