Question

Sol. [Key Concept: Telescoping: $\frac{k^{2 -}\frac{12}{}k^{4 +}\frac{14}{}}{}$ splits as a difference of consecutive terms.]

$$\sum_{k = 1}^{n}\mspace{2mu}\frac{k^{2} - \frac{1}{2}}{\left( k^{2} - k + \frac{1}{2} \right)\left( k^{2} + k + \frac{1}{2} \right)} = \sum_{k = 1}^{n}\mspace{2mu}\left( \frac{k - \frac{1}{2}}{k^{2} - k + \frac{1}{2}} - \frac{k + \frac{1}{2}}{k^{2} + k + \frac{1}{2}} \right)$$

$$= 1 - \frac{n + \frac{1}{2}}{n^{2} + n + \frac{1}{2}} = 1 - \frac{2n + 1}{2n^{2} + 2n + 1}$$