Question

[holicuoi]

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

[Trần Thị Loan]

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

<=> \(\frac{\left(2.3.4....n\right)^2}{\left(1.2.3...\left(n-1\right)\right).\left(3.4...\left(n+1\right)\right)}=\frac{2015}{1008}\)

<=> \(\frac{\left(2.3.4....n\right).\left(2.3.4....n\right)}{\left(1.2.3...\left(n-1\right)\right).\left(3.4...\left(n+1\right)\right)}=\frac{2015}{1008}\)

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

<=> 2n.1008 = 2015.(n+1)

<=> 2016n = 2015n + 2015 

<=> n = 2015

Answer 2

[Cố lên Tân]

\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{\left(n-1\right)\left(n+1\right)}\right)=1\frac{1007}{1008}=\left(1+\frac{1}{1.3}+\frac{1}{2.4}\right)=2.185897436\)