Bài này bạn nhân 2B ra

Sau đó tách mỗi phân số thành 2 hiệu

Từ đó triệt tiêu sẽ ra 2 số cuối cùng

Bạn trừ 2 số đó với nhau là ra

P/s : Sorry mình đg ôn thi nên ko nên giải trực tiếp, thông cảm nha

\(B=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{16}{2.51}=\frac{8}{51}\)

.

B = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

21176

Bạn tham khảo nhé ! ( Bạn chỉ việc đổi số thôi )

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.

Vậy A=172650

Ai muốn tham gia team thì kết bạn nhé , team tớ đang cần sự hỗ trợ từ các bạn ,STUDY WELL !

\(1\cdot3+3\cdot5+5\cdot7+...+49\cdot51=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{50}\right)=\frac{1}{2}\cdot\frac{49}{50}\)

\(=\frac{49}{100}\)

Chúc bạn học tốt ^^!!!

42875

Cho xin cách lam

1.3+2.4+ 3.5 +.........+ 48.50 + 49.51 =  [ 1+2+3+...............+49  ] x  [ 3+4+5.............+51 ]  =   1225  x 1323 =1620675

D = \(\frac{3}{1.3}\) + \(\frac{3}{3.5}\)+.......+ \(\frac{3}{49.51}\)

D = \(\frac{3.2}{1.3.2}\)+ \(\frac{3.2}{3.5.2}\)+ .....+ \(\frac{3.2}{49.51.2}\)

D = \(\frac{3}{2}\)( \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+...+ \(\frac{2}{49.51}\))

D = \(\frac{3}{2}\)( \(\frac{1}{1}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{5}\)+ .....+ \(\frac{1}{49}\)- \(\frac{1}{51}\))

D = \(\frac{3}{2}\)( 1 - \(\frac{1}{51}\))

D = \(\frac{3}{2}\) x  \(\frac{50}{51}\)

D = \(\frac{25}{17}\)

Vậy D = \(\frac{25}{17}\)

**** xcho mình nha bn !!!! 

tại sao lạ là 3/2 hả bạn

S = 1.3+2.4+3.5+ .....+48.50+49.51 

S = 1(2 + 1) + 2(3 + 1) + 3(4 + 1) + ... + 48(49 + 1) + 49(50 + 1)

S =1 + 1.2 + 2.3 + 2 + 3.4 + 3 + ... + 48.49 + 48 + 49.50 + 49

S = (1 + 2 + 3 + ... + 49) + (1.2 + 2.3 + 3.4 + ... + 49.50)

đặt A  = 1 + 2 + 3 + ... + 49

A = (49 + 1).49 : 2

A = 1225

đătj B = 1.2 + 2.3 + 3.4 + ... + 49.50

3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3B = 49.50.51

B = 41650

=> S = A + B = 41650 + 1225 = 42875

cảm ơn bạn nha Tomoe !!!

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(=\frac{99}{100}\)

b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=\frac{1}{3}-\frac{1}{51}\)

\(=\frac{16}{51}\)

a) 1/1.2+1/2.3+1/3.4+...+1/99.100

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/99 - 1/100

= 1/1 - 1/100

= 99/100

b) 2/3.5+2/5.7+...+2/49.51

= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )

= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )

= 2 . ( 1/3 - 1/50 )

= 2 . 47/150

= 47/75

                           \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

                 \(=\frac{2}{3}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

                  \(=\frac{2}{3}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

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

                  \(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)

              Ủng hộ mk nha !!! ^_^

25/17 mới đúng

3/1.3 + 3/3.5 + 3/5.7 + ....... + 3/49.51

= 3 x ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/49.51 )

= 3 x ( 1 - 1/51 )

= 3 x      50/51

=       150/151

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

 
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

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

\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{49.51}\)

= 3. \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{49}-\frac{1}{51}\right)\)

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

=\(\frac{3}{2}.\frac{16}{51}\)

=\(\frac{8}{17}\)

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