Question

Tìm n biết \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{\left(n-1\right)\left(n+1\right)}\right)=1\frac{1007}{1008}\)

Answer 1

=> \(\frac{4}{1.3}.\frac{9}{2.4}...\frac{n^2}{\left(n-1\right)\left(n+1\right)}=\frac{2015}{1008}\)

<=> \(\frac{2^2.3^2...n^2}{1.3.2.4....\left(n-1\right).\left(n+1\right)}=\frac{2015}{1008}\)

<=> \(\frac{\left(2.3.4....n\right).\left(2.3.4...n\right)}{\left(1.2.3...\left(n-1\right)\right).\left(3.4.5...\left(n+1\right)\right)}=\frac{2015}{1008}\)

<=> \(\frac{n.2}{n+1}=\frac{2015}{1008}\)

=> 1008.2n = 2015.(n+1)

<=> 2016n = 2015n + 2015

<=> n = 2015

*) Bạn hỏi câu này một lần rồi!!!

Answer 2

nhung hinh nhu ban lam sai de roi thi phai