Why so

By using Sophie Germain Identity ,

$T_n = n^{4} + \frac{1}{4} = [n(n+1) + \frac{1}{2} ][n(n-1) + \frac{1}{2} ]$

$T_{n-1} = [n(n-1) + \frac{1}{2} ][(n-1)(n-2) + \frac{1}{2} ]$

$\rightarrow \dfrac{T_n}{T_{n-1}} = \frac{n(n+1) + \frac{1}{2}}{(n-1)(n-2) + \frac{1}{2}}$

Product is $\dfrac{T_2}{T_1} \cdot \dfrac{T_4}{T_3} \cdot \dfrac{T_6}{T_5} \cdot \cdot \cdot \dfrac{T_{12}}{T_{11}}$

Again $\dfrac{T_n}{T_{n-1}} \cdot \dfrac{T_{n+2}}{T_{n+1}} = \frac{n(n+1) + \frac{1}{2}}{(n-1)(n-2) + \frac{1}{2}} \cdot \frac{(n+2)(n+3) + \frac{1}{2}}{n(n+1) + \frac{1}{2}} = \dfrac{T_{n+2}}{T_{n-1}}$

$\rightarrow \dfrac{T_2}{T_1} \cdot \dfrac{T_4}{T_3} \cdot \dfrac{T_6}{T_5} \cdot \cdot \cdot \dfrac{T_{12}}{T_{11}} = \frac{T_{12}}{T_1} = \dfrac{156 + \frac{1}{2}}{\frac{1}{2}} = 313$

I posted it earlier

I don,t understand why, T(n+1) = n (n+1) + 1/2 , where T(n) = n (n+1) + 1/2 ?

Hi sis, I request you to copy the LaTex from my comment and repost it yourself .

Sorry if I was intrusive but the quality of the image was not good enough, so I had to do it .

Please bear with me $\ddot\smile$ .