B=22/1.2.32/2.4.42/3.5...502/49.51
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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
\(B=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{16}{2.51}=\frac{8}{51}\)
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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 !
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 !!!!
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
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 !!! ^_^
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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