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24 tháng 3 2019

Ta có:

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+........+\frac{2500}{49.51}\)

1 tháng 2 2020

\(S=\frac{4}{1\times3}+\frac{16}{3\times5}+\frac{36}{5\times7}+...+\frac{2500}{49\times51}\)

\(=\frac{1\times3+1}{1\times3}+\frac{3\times5+1}{3\times5}+\frac{5\times7+1}{5\times7}+...+\frac{49\times51+1}{49\times51}\)

\(=\frac{1\times3}{1\times3}+\frac{1}{1\times3}+\frac{3\times5}{3\times5}+\frac{1}{3\times5}+\frac{5\times7}{5\times7}+\frac{1}{5\times7}+...+\frac{49\times51}{49\times51}+\frac{1}{49\times51}\)

\(=1+\frac{1}{1\times3}+1+\frac{1}{3\times5}+1+\frac{1}{5\times7}+...+\frac{1}{49\times51}\) (  Có : \(\left(51-3\right)\div2+1=25\)chữ số 1 )

\(=25+\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{49\times51}\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\frac{50}{51}\)

\(=25+\frac{25}{51}\)

\(=\frac{1300}{51}\)

1 tháng 2 2020

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)

\(=\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+...+\frac{2500}{2499}\)

\(=1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+...+1+\frac{1}{2499}\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2500}\right)\)

\(=25+\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

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

\(=1-\frac{1}{51}=\frac{50}{51}\)

\(\Rightarrow S=25+\frac{50}{51}=\frac{1325}{51}\)

Vậy S=\(\frac{1325}{51}\)

AH
Akai Haruma
Giáo viên
6 tháng 12 2023

Lời giải:

$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$

$=1-\frac{1}{99}=\frac{98}{99}$

Ta có: \(S=\dfrac{4}{1\cdot3}+\dfrac{16}{3\cdot5}+\dfrac{36}{5\cdot7}+...+\dfrac{2500}{49\cdot51}\)

\(=1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{3\cdot5}+1+\dfrac{1}{5\cdot7}+...+1+\dfrac{1}{49\cdot51}\)

\(=25+\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)

\(=25+\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=25+\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)

\(=25+\dfrac{1}{2}\cdot\dfrac{50}{51}\)

\(=25+\dfrac{25}{51}\)

\(=25\cdot\dfrac{52}{51}=\dfrac{1300}{51}\)

30 tháng 1 2023

sai gòi

 

 

AH
Akai Haruma
Giáo viên
12 tháng 8 2023

Lời giải:

Gọi tổng trên là $A$
$A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)$

$=2\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{25}-\frac{1}{27}\right)$

$=2\left(1-\frac{1}{27})=\frac{52}{27}$

25 tháng 7 2016

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

24 tháng 6 2017

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(M=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(M=2\left(1-\frac{1}{100}\right)\)

\(M=2.\frac{99}{100}\)

\(M=\frac{99}{50}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)

\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\frac{98}{99}\)

\(N=\frac{49}{33}\)