Question:

Step 1: Group the terms in pairs. We rewrite the sum $S$ by pairing consecutive terms: $$S = (a_1^2 - a_2^2) + (a_3^2 - a_4^2) + \cdots + (a_{2k-1}^2 - a_{2k}^2)$$ This gives us $k$ pairs in total. Step 2: Factor each pair using the difference of squares formula. For each pair, we apply the algebraic identity $x^2 - y^2 = (x-y)(x+y)$: $$a_{2r-1}^2 - a_{2r}^2 = (a_{2r-1} - a_{2r})(a_{2r-1} + a_{2r})$$ Since the terms are in arithmetic progression with common difference $d$, we have: $$a_{2r} - a_{2r-1} = d$$ Therefore: $$a_{2r-1} - a_{2r} = -d$$ So each pair becomes: $$a_{2r-1}^2 - a_{2r}^2 = -d(a_{2r-1} + a_{2r})$$ Step 3: Sum all the pairs. $$S = -d \sum_{r=1}^{k} (a_{2r-1} + a_{2r})$$ The sum inside can be rewritten as: $$\sum_{r=1}^{k} (a_{2r-1} + a_{2r}) = a_1 + a_2 + a_3 + a_4 + \cdots + a_{2k-1} + a_{2k}$$ This is the sum of the first $2k$ terms of an arithmetic progression. Step 4: Apply the arithmetic progression sum formula. For an AP, the sum of $n$ terms is $\frac{n}{2}(\text{first term} + \text{last term})$. Therefore: $$a_1 + a_2 + \cdots + a_{2k} = \frac{2k}{2}(a_1 + a_{2k}) = k(a_1 + a_{2k})$$ Step 5: Express the common difference in terms of $a_1$ and $a_{2k}$. The $2k$-th term of an AP is given by: $$a_{2k} = a_1 + (2k-1)d$$ Solving for $d$: $$d = \frac{a_{2k} - a_1}{2k-1}$$ Step 6: Substitute back to find $S$. $$S = -d \cdot k(a_1 + a_{2k}) = -\frac{a_{2k} - a_1}{2k-1} \cdot k(a_1 + a_{2k})$$ $$S = \frac{a_1 - a_{2k}}{2k-1} \cdot k(a_1 + a_{2k})$$ $$S = \frac{k(a_1 - a_{2k})(a_1 + a_{2k})}{2k-1}$$ Step 7: Apply the difference of squares formula to simplify. Using $x^2 - y^2 = (x-y)(x+y)$: $$(a_1 - a_{2k})(a_1 + a_{2k}) = a_1^2 - a_{2k}^2$$ Therefore: $$S = \frac{k(a_1^2 - a_{2k}^2)}{2k-1} = \frac{k}{2k-1}(a_1^2 - a_{2k}^2)$$ **Final Answer:** The expression for $S$ is $\boxed{\frac{k}{2k-1}(a_1^2 - a_{2k}^2)}$, which matches **Option 1**.