Question

Let $T_1=T_2=1$, where $T_1,T_2,T_3,...,T_n$ is a sequence and $T_{n+2}=(T_{n+1}{)}^{-1}+T_n$; $n=1,2,3$,... Find $T_{2019}$.

• A. $\frac{\left(2019\right)\times \left(2017\right)\times \left(2015\right)\times ....\times \left(1\right)}{\left(2018\right)\times \left(2016\right)\times \left(2014\right)\times ....\times \left(2\right)}$
• B. $\frac{\left(2018\right)\times \left(2016\right)\times \left(2014\right)\times ....\times \left(2\right)}{\left(2017\right)\times \left(2015\right)\times \left(2013\right)\times ....\times \left(1\right)}$
• C. $\frac{\left(2018\right)\times \left(2016\right)\times \left(2014\right)\times ....\times \left(2\right)}{\left(2019\right)\times \left(2017\right)\times \left(2015\right)\times ....\times \left(1\right)}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\left(2018\right)\times \left(2016\right)\times \left(2014\right)\times ....\times \left(2\right)}{\left(2017\right)\times \left(2015\right)\times \left(2013\right)\times ....\times \left(1\right)}$

$T_{n+2}=T_n+\frac{1}{T_{n+1}}$
$\Rightarrow T_{n+2}-T_n=\frac{1}{T_{n+1}}$
$\Rightarrow T_{n+2}\cdot T_{n+1}-T_{n+1}\cdot T_n=1$

Now, let $a_n=T_{n+1}\cdot T_n$
$\Rightarrow a_{n+1}-a_n=1$
So, $a_1,a_2,a_3,...,a_{n+1}$ form an A.P. with common difference, $d=1$; and first term $a_1=T_2\cdot T_1=\left(1\right)\left(1\right)=1$
Now, $n^{th}$ term of this AP is $a_n=a_1+(n-1)d$
$a_n=1+(n-1)1$
$a_n=n$
So, $T_{n+1}\cdot T_n=n$
So, $\left(T_{2019}\right)\left(T_{2018}\right)=2018$ and $\left(T_{2018}\right)\left(T_{2017}\right)=2017$

$\Rightarrow T_{2019}=\frac{2018}{T_{2018}}=\frac{2018}{\left(\frac{2017}{T_{2017}}\right)}-\left(\frac{2018}{2017}\right)T_{2017}$

Thus, $T_{2019}=\left(\frac{2018}{2017}\right)T_{2017}$

Similarly, $T_{2017}=\left(\frac{2016}{2015}\right)T_{2015}$

$T_{2015}=\left(\frac{2014}{2013}\right)T_{2013}$

$T_3=\left(\frac{2}{1}\right)T_2=2$

Thus, $T_{2019}=\left(\frac{2018}{2017}\right)\left(\frac{2016}{2015}\right)\left(\frac{2014}{2013}\right)\left(\frac{2}{1}\right)$

So $T_{2019}=\frac{\left(2018\right)\left(2016\right)\left(2014\right)\left(2\right)}{\left(2017\right)\left(2015\right)\left(2013\right).\left(3\right)\left(1\right)}$
So option B.