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TH
1
10 tháng 4 2017
C=1/15+1/35+1/63+..+1/2499
=1/3.5+1/5.7+1/7.9+...+1/49.51
=1/2(2/3.5+2/5.7+2/7.9+...+2/49.51)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/49-1/51)
= 1/2.(1/3-1/51)
=1/2.16/51
=8/51
TH
3
6 tháng 4 2017
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
6 tháng 4 2017
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)
26 tháng 4 2018
Câu \(C=\left(...\right)\) thiếu đề
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(............\)
\(\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow\)\(D=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(\Rightarrow\)\(D< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\)\(D< 1-\frac{1}{10}< 1\)
\(\Rightarrow\)\(D< 1\) ( đpcm )
Vậy \(D< 1\)
Chúc bạn học tốt ~
26 tháng 4 2018
\(C=\frac{1}{15}+\frac{1}{35}+....+\frac{1}{2499}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+........+\frac{1}{49.51}\)
\(C=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\frac{16}{51}\)
\(C=\frac{8}{51}\)
\(D=\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{10^2}\)
ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(....................\)
\(\frac{1}{10^2}=\frac{1}{10.10}< \frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{9.10}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow D< 1-\frac{1}{10}\)
\(\Rightarrow D< \frac{9}{10}\) ( 1 )
mà \(\frac{9}{10}< 1\) ( 2 )
từ ( 1 ) và ( 2 ) \(\Rightarrow D< 1\left(\text{đ}pcm\right)\)
SN
4
WH
6 tháng 5 2018
Trả lời
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow C=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{...1}{49\cdot51}\)
\(\Rightarrow2C=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\right)\)
\(\Rightarrow2C=\frac{2}{1\cdot3}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(\Rightarrow2C=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2C=1-\frac{1}{51}\)
\(\Rightarrow2C=\frac{50}{51}\)
\(\Rightarrow C=\frac{50}{51}:2\)
\(\Rightarrow C=\frac{25}{51}\)
Vậy \(C=\frac{25}{51}\)
AM
4 tháng 7 2015
a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)
5 tháng 5 2017
Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha
b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{3}-\frac{1}{51}\)
\(2A=\frac{16}{51}\)
\(A=\frac{16}{51}:2\)
\(A=\frac{8}{51}\)
LT
5
B
26 tháng 4 2018
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)
\(=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
26 tháng 4 2018
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(C=\frac{1}{3}-\frac{1}{51}\)
\(C=\frac{16}{51}\)
24 tháng 2 2020
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{16}{15}\right)...\left(1+\frac{2500}{2499}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{2500}{2499}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{50.50}{49.51}\)
\(=2.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}...\frac{50}{49}.\frac{50}{51}\)
\(=\left(2.\frac{3}{2}.\frac{4}{3}...\frac{50}{49}\right)\left(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{50}{51}\right)\)
\(=\frac{2.3.4...50}{2.3.4...49}.\frac{2.3.4...50}{3.4.5...51}\)
\(=50.\frac{2}{51}=\frac{100}{51}\)
7 tháng 6 2016
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
Ta có :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{2499}\right)\) ( bước này hơi khó hiểu tí nhé )
\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\) ( phân tích mẫu )
\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\) ( áp dụng công thức thoi )
\(=\)\(\frac{1}{2}\left(1-\frac{1}{51}\right)\) ( loại bỏ nhưng phân số đối nhau )
\(=\)\(\frac{1}{2}.\frac{50}{51}\)
\(=\)\(\frac{25}{51}\)
Chúc bạn học tốt ~
ĐẶT \(A\)\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(2.A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(2.A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2.A=1-\frac{1}{51}\)
\(2.A=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}\div2=\frac{25}{51}\)