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=> \(\frac{4}{1.3}.\frac{9}{2.4}...\frac{n^2}{\left(n-1\right)\left(n+1\right)}=\frac{2015}{1008}\)
<=> \(\frac{2^2.3^2...n^2}{1.3.2.4....\left(n-1\right).\left(n+1\right)}=\frac{2015}{1008}\)
<=> \(\frac{\left(2.3.4....n\right).\left(2.3.4...n\right)}{\left(1.2.3...\left(n-1\right)\right).\left(3.4.5...\left(n+1\right)\right)}=\frac{2015}{1008}\)
<=> \(\frac{n.2}{n+1}=\frac{2015}{1008}\)
=> 1008.2n = 2015.(n+1)
<=> 2016n = 2015n + 2015
<=> n = 2015
*) Bạn hỏi câu này một lần rồi!!!
bạn kiểm tra lại đề nhé! vì số hạng tổng quát chẳng liên quan gì đến số hạng đầu
Có thể đề đúng là: \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{\left(n-1\right)\left(n+1\right)}\right)=1\frac{1007}{1008}\)
Cậu có thể vào đây tham khảo : http://h.vn/hoi-dap/question/119685.html
\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{20.22}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{20.22+1}{20.22}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{21^2}{20.22}\)
\(=\frac{\left(2.3.4.....21\right)\left(2.3.4.....21\right)}{\left(1.2.3.....20\right)\left(3.4.5.....22\right)}\)
\(=\frac{21.2}{22}=\frac{42}{22}=\frac{21}{11}\)
\(S=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2016.2018}\right)\)
\(\Rightarrow S=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2016.2018+1}{2016.2018}\)
\(\Rightarrow S=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2017^2}{2016.2018}\)
\(\Rightarrow S=\frac{\left(2.3.4.....2017\right)\left(2.3.4.....2017\right)}{\left(1.2.3.....2016\right)\left(3.4.5.....2018\right)}\)
\(\Rightarrow S=\frac{2017.2}{1.2018}=\frac{4034}{2018}=\frac{2017}{1009}\)
<=> \(\frac{4}{1.3}.\frac{9}{2.4}...\frac{n^2}{\left(n-1\right)\left(n+1\right)}=\frac{2015}{1008}\)
<=> \(\frac{\left(2.3.4....n\right)^2}{\left(1.2.3...\left(n-1\right)\right).\left(3.4...\left(n+1\right)\right)}=\frac{2015}{1008}\)
<=> \(\frac{\left(2.3.4....n\right).\left(2.3.4....n\right)}{\left(1.2.3...\left(n-1\right)\right).\left(3.4...\left(n+1\right)\right)}=\frac{2015}{1008}\)
<=> \(\frac{n.2}{n+1}=\frac{2015}{1008}\)
<=> 2n.1008 = 2015.(n+1)
<=> 2016n = 2015n + 2015
<=> n = 2015
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{\left(n-1\right)\left(n+1\right)}\right)=1\frac{1007}{1008}=\left(1+\frac{1}{1.3}+\frac{1}{2.4}\right)=2.185897436\)