Find Sum of Series of Quadratic Squares Terms

To find the sum of the given series up to $10$ terms: $S_{10} = 1 + \frac{1}{2}(1^2 + 2^2) + \frac{1}{3}(1^2 + 2^2 + 3^2) + \dots \text{ upto } 10 \text{ terms}$

Let $T_n$ denote the $n$-th term of the series. The general term $T_n$ is given by: $T_n = \frac{1}{n} \sum_{k=1}^{n} k^2$

Using the formula for the sum of squares of the first $n$ natural numbers, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, we substitute this into the expression for $T_n$: $T_n = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6} = \frac{2n^2 + 3n + 1}{6}$

The sum of the first $10$ terms, $S_{10}$, is: $S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} \frac{2n^2 + 3n + 1}{6} = \frac{1}{6} \left[ 2 \sum_{n=1}^{10} n^2 + 3 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 1 \right]$

Now, we evaluate each of the summations for $n = 10$:

1. $\sum_{n=1}^{10} 1 = 10$
2. $\sum_{n=1}^{10} n = \frac{10 \times (10 + 1)}{2} = 55$
3. $\sum_{n=1}^{10} n^2 = \frac{10 \times (10 + 1) \times (2 \cdot 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 385$

Substituting these values back into the expression for $S_{10}$: $S_{10} = \frac{1}{6} \left[ 2(385) + 3(55) + 10 \right]$ $S_{10} = \frac{1}{6} \left[ 770 + 165 + 10 \right]$ $S_{10} = \frac{945}{6} = \frac{315}{2}$

Thus, the sum of the series up to 10 terms is $\frac{315}{2}$.