1. **Problem statement:** Calculate the value of the expression $$\frac{(2 \times 5 + 2)(4 \times 7 + 2)(6 \times 9 + 2) \cdots (1994 \times 1997 + 2)}{(1 \times 4 + 2)(3 \times 6 + 2)(5 \times 8 + 2) \cdots (1993 \times 1996 + 2)}$$ 2. **Understanding the pattern:** - Numerator terms are of the form $$(2k) \times (2k+3) + 2$$ for $k=1,2,\ldots,997$. - Denominator terms are of the form $$(2k-1) \times (2k+2) + 2$$ for $k=1,2,\ldots,997$. 3. **Simplify each term:** - Numerator term: $$ (2k)(2k+3) + 2 = 4k^2 + 6k + 2 $$ - Denominator term: $$ (2k-1)(2k+2) + 2 = (4k^2 + 4k - 2k - 2) + 2 = 4k^2 + 2k $$ 4. **Rewrite the fraction:** $$ \frac{\prod_{k=1}^{997} (4k^2 + 6k + 2)}{\prod_{k=1}^{997} (4k^2 + 2k)} $$ 5. **Factor numerator terms:** $$4k^2 + 6k + 2 = 2(2k^2 + 3k + 1) = 2(2k+1)(k+1)$$ 6. **Factor denominator terms:** $$4k^2 + 2k = 2k(2k+1)$$ 7. **Substitute back:** $$ \frac{\prod_{k=1}^{997} 2(2k+1)(k+1)}{\prod_{k=1}^{997} 2k(2k+1)} = \frac{\prod_{k=1}^{997} 2 \times (2k+1) \times (k+1)}{\prod_{k=1}^{997} 2 \times k \times (2k+1)} $$ 8. **Cancel common factors:** - The factor $2$ appears in numerator and denominator 997 times, cancels out. - The factor $(2k+1)$ appears in numerator and denominator for each $k$, cancels out. 9. **Remaining terms:** $$ \frac{\prod_{k=1}^{997} (k+1)}{\prod_{k=1}^{997} k} = \frac{2 \times 3 \times 4 \times \cdots \times 998}{1 \times 2 \times 3 \times \cdots \times 997} $$ 10. **Simplify the fraction:** - Numerator is $998! / 1$ - Denominator is $997!$ So, $$ \frac{998!}{997!} = 998 $$ **Final answer:** $$\boxed{998}$$ This corresponds to option B in the multiple-choice answers.