For arbitrary $a, k \in \mathbb{N}^+$, if $k^2 < a < (k+1)^2$, we set
$$a = k^2 + m, \quad m = 1, 2, \dots, 2k,$$
$$\sqrt{a} = k + \theta, \quad 0 < \theta < 1,$$
$$\left[ \frac{1}{\sqrt{a}} \right] = \left[ \frac{1}{\sqrt{a-k}} \right] = \left[ \frac{\sqrt{a+k}}{a-k^2} \right] = \left[ \frac{2k+\theta}{m} \right].$$
$$\text{For} \quad 0 < \frac{2k+\theta}{m} - \frac{2k}{\theta} < 1,$$
if there exists an integer $t$ between $\frac{2k}{m}$ and $\frac{2k+\theta}{m}$, then
$$\frac{2k}{m} < t < \frac{2k+\theta}{m}.$$
On one hand $2k < mt$, thus $2k+1 \le mt$. On the other hand, $mt < 2k+\theta < 2k+1$, a contradiction.
$$\text{Thus} \quad \left[ \frac{2k+\theta}{m} \right] = \left[ \frac{2k}{m} \right], \\ \sum_{k<a<k+1} \left[ \frac{1}{\{a\}} \right] = \sum_{m=1}^{2k} \left[ \frac{2k}{m} \right], \\ \sum_{a=1}^{n+1} f(a) = \sum_{k=1}^{n} \sum_{i=1}^{2k} \left[ \frac{2k}{i} \right]. \quad (1)$$

Thus
$$\begin{aligned} \sum_{a=1}^{(n+1)^2} f(a) &= \sum_{k=1}^{n} \sum_{j=1}^{2k} T(j) \\ &= n[T(1) + T(2)] + (n-1)[T(3) + T(4)] \\ &\quad + \dots + [T(2n-1) + T(2n)] \end{aligned} \quad (2)$$
$$\text{From (2), } \sum_{k=1}^{16^2} f(k) = \sum_{k=1}^{15} (16-k)[T(2k-1) + T(2k)]. \quad (3)$$