Key Concepts and Formulas

• Sum of the first $k$ natural numbers: The sum of an arithmetic progression $1 + 2 + 3 + \dots + k$ is given by: $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$
• Sum of the squares of the first $n$ natural numbers: The sum of the squares of the first $n$ natural numbers is given by: $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Step-by-Step Solution

Step 1: Simplify the expression for $S_k$ We are given $S_k = \frac{1 + 2 + 3 + \dots + k}{k}$. Why this step? Simplifying the general term $S_k$ is essential to make the subsequent calculations manageable. It transforms a complex expression into a more straightforward algebraic form. Using the formula for the sum of the first $k$ natural numbers, we have: $S_k = \frac{\frac{k(k+1)}{2}}{k}$ For $k \ge 1$, we can cancel $k$ from the numerator and denominator: $S_k = \frac{k+1}{2}$

Step 2: Find the expression for $S_k^2$ The problem requires the sum of the squares of $S_k$, so we need to square the simplified expression for $S_k$. Why this step? This step directly prepares the term $S_k^2$ for summation, as required by the problem statement. Squaring $S_k$: $S_k^2 = \left(\frac{k+1}{2}\right)^2 = \frac{(k+1)^2}{4}$

Step 3: Evaluate the summation $\sum_{k=1}^{10} S_k^2$ We need to compute the sum $S_1^2 + S_2^2 + \dots + S_{10}^2$. Why this step? This is the core computational part of the problem where we apply the summation formulas to find the value of the given series. $\sum_{k=1}^{10} S_k^2 = \sum_{k=1}^{10} \frac{(k+1)^2}{4}$ We can factor out the constant $\frac{1}{4}$: $\sum_{k=1}^{10} S_k^2 = \frac{1}{4} \sum_{k=1}^{10} (k+1)^2$ To evaluate $\sum_{k=1}^{10} (k+1)^2$, we can use a change of index. Let $j = k+1$. When $k=1$, $j=2$. When $k=10$, $j=11$. So, the summation becomes: $\sum_{k=1}^{10} (k+1)^2 = \sum_{j=2}^{11} j^2$ This is the sum of squares from $2^2$ to $11^2$. We can express this using the formula for the sum of the first $n$ squares: $\sum_{j=2}^{11} j^2 = \left(\sum_{j=1}^{11} j^2\right) - 1^2$ Using the formula $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ with $n=11$: $\sum_{j=1}^{11} j^2 = \frac{11(11+1)(2 \cdot 11+1)}{6} = \frac{11 \cdot 12 \cdot 23}{6} = 11 \cdot 2 \cdot 23 = 506$ Therefore, $\sum_{j=2}^{11} j^2 = 506 - 1^2 = 506 - 1 = 505$ Substituting this back into the expression for $\sum_{k=1}^{10} S_k^2$: $\sum_{k=1}^{10} S_k^2 = \frac{1}{4} \cdot 505 = \frac{505}{4}$

Step 4: Solve for A We are given that $S_1^2 + S_2^2 + \dots + S_{10}^2 = \frac{5}{12}A$. Why this step? This step involves equating our calculated sum with the given expression involving $A$ to isolate and solve for the unknown variable $A$. We have calculated the sum to be $\frac{505}{4}$. So, we set up the equation: $\frac{505}{4} = \frac{5}{12}A$ To solve for $A$, we multiply both sides by $\frac{12}{5}$: $A = \frac{505}{4} \cdot \frac{12}{5}$ Simplify the expression: $A = \frac{505}{5} \cdot \frac{12}{4} = 101 \cdot 3 = 303$

Common Mistakes & Tips

• Index of summation: Be careful when changing the index of summation. Ensure the new limits (lower and upper bounds) are correctly adjusted.
• Arithmetic errors: Summation formulas involve multiplications and divisions. Double-check all arithmetic calculations to avoid errors that can lead to an incorrect final answer.
• Formula recall: Ensure accurate recall of standard summation formulas for natural numbers and their squares.

Summary

The problem involves calculating the sum of squares of the average of the first $k$ natural numbers. We first simplified the expression for $S_k$ to $\frac{k+1}{2}$. Then, we squared this to get $S_k^2 = \frac{(k+1)^2}{4}$. By evaluating the summation $\sum_{k=1}^{10} S_k^2$ using the formula for the sum of squares and a change of index, we found the sum to be $\frac{505}{4}$. Finally, by equating this sum to the given expression $\frac{5}{12}A$, we solved for $A$ and found it to be 303.

The final answer is $\boxed{303}$ which corresponds to option (D).