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\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)
\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)
\(=\frac{2014\cdot2}{1\cdot2015}\)
\(=\frac{4028}{2015}\)
\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)
\(Q=\frac{1}{100}\)
\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)
\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)
\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)
Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới
\(P=\frac{201}{100}\)
=1/2.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)
=1/2.(1+1/1-1/3).(1+1/3-1/5)...(1+1/2014-1/2016)
=1/2.1+(1/1-1/2016)
=1/2.2015/2016
=2015/4032
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(A=\frac{2.3.4...2015}{1.2.3...2014}.\frac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\frac{1}{1008}\)
\(A=\frac{2015}{1008}\)
Ta có công thức : với n thuộc N* thì ta luôn có :
\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Áp dụng vào bài toán ta được :
\(P=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{49.51}\right)+\frac{2}{51}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}+\frac{2}{51}\)
\(=\frac{\left(2.3.4...50\right)\left(2.3.4...50\right)}{\left(1.2.3...49\right)\left(3.4.5....51\right)}+\frac{2}{51}\)
\(=\frac{50.2}{51}+\frac{2}{51}=\frac{102}{51}=2\)