If ( n^r - tn + 1/4 ) be a perfect square, then the values of t are = ?

Answer: Option D

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