$${\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right)$$     be a perfect square, then the values of t are = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr} $$