1. **Stating the problem:** We have two products to analyze and compare: - Product A: $(n+1)(n+2)(n+3)\ldots(n+2s-1)(n+2s)$ - Product B: $2 \cdot 4 \cdot 6 \cdots (2s-2)(2s) \times (2n+3)(2n+5) \cdots (2n+2s+1)$ We want to understand their relationship or simplify if possible. 2. **Rewrite Product A:** Product A contains $2s$ terms starting at $(n+1)$ and increasing by 1 up to $(n+2s)$. This can be expressed using factorials as $$ \frac{(n+2s)!}{n!} $$ because $\frac{(n+2s)!}{n!} = (n+1)(n+2)\cdots (n+2s)$. 3. **Rewrite Product B:** Product B consists of two parts: - Even numbers from 2 to $2s$, which is $2 \cdot 4 \cdot 6 \cdots 2s$. - Then odd numbers starting from $(2n+3)$ to $(2n+2s+1)$ with increment 2. The product of even numbers can be written using factorial: $$ 2 \cdot 4 \cdot 6 \cdots (2s) = 2^{s} \cdot (1 \cdot 2 \cdot 3 \cdots s) = 2^{s} s! $$ The product of odd numbers is: $$ (2n+3)(2n+5) \cdots (2n+2s+1) $$ There are $s$ terms, each increasing by 2. 4. **Conclusion and identification:** - Product A is $\frac{(n+2s)!}{n!}$. - Product B is $ 2^{s} s! \times \prod_{k=1}^s [2n + (2k + 1)]$ which matches with the product of odd numbers part. This suggests a factorization approach where product A equals product B if the odd numbers part corresponds to $\frac{(n+2s)!}{n! \cdot 2^{s} s!}$. Recognizing this relates to combinatorial identities involving double factorials. 5. **Summary:** The first product is a simple factorial ratio; the second is a product of even factorial term $2^{s} s!$ times a product over odd terms starting at $(2n+3)$. If we wanted to exactly equate or simplify further additional context or values for $n$ and $s$ are needed. **Final simplified forms:** $$ \text{Product A} = \frac{(n+2s)!}{n!} $$ $$ \text{Product B} = 2^{s} s! \times \prod_{k=1}^s (2n + 2k + 1) $$