The sum $$1 + \dfrac{1}{2}(1^2 + 2^2) + \dfrac{1}{3}(1^2 + 2^2 + 3^2) + \ldots$$ upto 10 terms is equal to :

$$T_k = \frac{1}{k} \left( 1^2 + 2^2 + 3^2 + \dots + k^2 \right)$$

$$T_k = \frac{1}{k} \cdot \left[ \frac{k(k+1)(2k+1)}{6} \right]$$

$$T_k = \frac{(k+1)(2k+1)}{6}$$

$$T_k = \frac{2k^2 + k + 2k + 1}{6} = \frac{2k^2 + 3k + 1}{6}$$

$$S_{10} = \sum_{k=1}^{10} T_k = \sum_{k=1}^{10} \frac{2k^2 + 3k + 1}{6}$$

$$S_{10} = \frac{1}{6} \left[ 2\sum_{k=1}^{10} k^2 + 3\sum_{k=1}^{10} k + \sum_{k=1}^{10} 1 \right]$$

$$\sum_{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} = 385$$

$$\sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55$$

$$\sum_{k=1}^{10} 1 = 10$$

$$S_{10} = \frac{1}{6} \left[ 2(385) + 3(55) + 10 \right]$$

$$= \frac{315}{2}$$