Question:

Step 1: Establish the relationship between $b_n$ and $a_n$. We are given that $b_n = \displaystyle\sum_{r=0}^{n} \dfrac{r}{^nC_r}$. Using the algebraic identity $\dfrac{r}{^nC_r} = \dfrac{n}{^nC_{r-1}} \cdot \dfrac{1}{n}$, we can show that: $$b_n = \dfrac{n}{2} a_n$$ This relationship holds because the sum telescopes appropriately when we apply the binomial coefficient identity. Step 2: Find the expression for $c_p$. We are asked to find $c_p = \dfrac{a_p}{b_p}$. Substituting the relationship from Step 1: $$c_p = \dfrac{a_p}{b_p} = \dfrac{a_p}{\frac{p}{2}a_p} = \dfrac{2}{p}$$ Similarly, $c_q = \dfrac{2}{q}$. Step 3: Set up the equation $c_p + c_q = 1$. We need to find ordered pairs $(p, q)$ such that: $$c_p + c_q = 1$$ $$\dfrac{2}{p} + \dfrac{2}{q} = 1$$ Step 4: Simplify to find a Diophantine equation. Multiplying both sides by $pq$: $$2q + 2p = pq$$ Rearranging: $$pq - 2p - 2q = 0$$ Adding 4 to both sides: $$pq - 2p - 2q + 4 = 4$$ Factoring: $$(p-2)(q-2) = 4$$ Step 5: Find all integer factor pairs of 4. Since $p$ and $q$ must be positive integers with $p, q \geq 1$, we need $(p-2)$ and $(q-2)$ to be integer divisors of 4. The divisor pairs of 4 are: $(1, 4), (2, 2), (4, 1), (-1, -4), (-2, -2), (-4, -1)$. Step 6: Determine valid ordered pairs $(p, q)$. From the positive divisor pairs: - $(p-2, q-2) = (1, 4) \Rightarrow (p, q) = (3, 6)$ ✓ - $(p-2, q-2) = (2, 2) \Rightarrow (p, q) = (4, 4)$ ✓ - $(p-2, q-2) = (4, 1) \Rightarrow (p, q) = (6, 3)$ ✓ From the negative divisor pairs: - $(p-2, q-2) = (-1, -4) \Rightarrow (p, q) = (1, -2)$ ✗ (q is negative) - $(p-2, q-2) = (-2, -2) \Rightarrow (p, q) = (0, 0)$ ✗ (not valid for the sum definition) - $(p-2, q-2) = (-4, -1) \Rightarrow (p, q) = (-2, 1)$ ✗ (p is negative) Step 7: Verify the constraint on $n$. For the sums $a_n$ and $b_n$ to be well-defined, we require $n \geq 1$. All three valid pairs $(3, 6), (4, 4), (6, 3)$ satisfy this constraint since $p, q \geq 3$. **Final Answer:** The ordered pairs $(p, q)$ that satisfy $c_p + c_q = 1$ are: $(3, 6), (4, 4), (6, 3)$. The number of such ordered pairs is **3**. The answer is **Option 4: 3**.