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\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n.\left(2n+2\right)}\))
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{4}-\frac{1}{2.\left(2n+2\right)}\)
\(=\frac{1}{4}-\frac{1}{4n+4}=\frac{1}{4}-\frac{1}{4.\left(n+1\right)}\)
\(=\frac{n+1}{4.\left(n+1\right)}-\frac{1}{4.\left(n+1\right)}=\frac{n+1-1}{4.\left(n+1\right)}=\frac{n}{4.\left(n+1\right)}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2A=1-\frac{1}{2n+1}\)
\(A=\frac{1}{2}-\frac{1}{\left(2n+1\right).2}< \frac{1}{2}\)
Vậy:...
- Hok tốt ~
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}+\frac{1}{2n+1}\)
=>\(2A=1-\frac{1}{2n-1}\)
=>\(2A=\frac{2n}{2n+1}\)
=>\(A=\frac{2n}{4n+2}=\frac{2n}{2\left(n+1\right)}=\frac{n}{n+1}< \frac{1}{2}\)
zậy A<1/2
A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)
tự làm là hạnh phúc của mỗi công dân.