Key Concepts and Formulas

1. General Term of a Series: Identifying a formula for the $r^{th}$ term ($T_r$) of a series is crucial for summation.
2. Summation Notation: The series can be represented concisely using $\sum_{r=1}^{n} T_r$.
3. Linearity of Summation: $\sum_{r=1}^{n} (ca_r \pm db_r) = c\sum_{r=1}^{n} a_r \pm d\sum_{r=1}^{n} b_r$ for constants $c, d$.
4. Standard Summation Formulas:
   • $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$
   • $\sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2$
   • $\sum_{r=1}^{n} c = cn$

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Step-by-Step Solution

The given series is $S = 1 + (1 - 2^2 \cdot 1) + (1 - 4^2 \cdot 3) + (1 - 6^2 \cdot 5) + \ldots + (1 - 20^2 \cdot 19)$. We are given that $S = \alpha - 220\beta$.

Step 1: Identify the General Term ($T_r$) and the Range of Summation. Let's analyze the terms within the parentheses. They follow the pattern $(1 - (\text{even number})^2 \cdot (\text{odd number}))$. The even numbers are $2, 4, 6, \ldots, 20$. If we let the $r^{th}$ even number be $2r$, then the corresponding odd number is $2r-1$. So, the general term inside the parentheses can be written as $T_r = 1 - (2r)^2 (2r-1)$.

Let's check the values of $r$: For $r=1$: $1 - (2 \cdot 1)^2 (2 \cdot 1 - 1) = 1 - 2^2 \cdot 1$. This is the first term in parentheses. For $r=2$: $1 - (2 \cdot 2)^2 (2 \cdot 2 - 1) = 1 - 4^2 \cdot 3$. This is the second term in parentheses. For the last term $(1 - 20^2 \cdot 19)$, we have $2r = 20$, which means $r = 10$. Thus, the terms in parentheses are generated by $r$ from $1$ to $10$.

The initial term '1' does not fit this pattern directly and is treated separately. Therefore, the sum can be written as: $S = 1 + \sum_{r=1}^{10} [1 - (2r)^2 (2r-1)]$

Step 2: Simplify the General Term. We expand and simplify the expression for $T_r$: $T_r = 1 - (2r)^2 (2r-1)$ $T_r = 1 - (4r^2)(2r-1)$ $T_r = 1 - (8r^3 - 4r^2)$ $T_r = 1 - 8r^3 + 4r^2$

Step 3: Apply Summation Properties and Formulas. Now, we substitute the simplified $T_r$ back into the sum: $S = 1 + \sum_{r=1}^{10} (1 - 8r^3 + 4r^2)$ Using the linearity of summation: $S = 1 + \sum_{r=1}^{10} 1 - \sum_{r=1}^{10} 8r^3 + \sum_{r=1}^{10} 4r^2$ $S = 1 + (1 \cdot 10) - 8\sum_{r=1}^{10} r^3 + 4\sum_{r=1}^{10} r^2$ $S = 1 + 10 - 8\sum_{r=1}^{10} r^3 + 4\sum_{r=1}^{10} r^2$ $S = 11 - 8\sum_{r=1}^{10} r^3 + 4\sum_{r=1}^{10} r^2$

Now, we use the standard summation formulas with $n=10$: $\sum_{r=1}^{10} r^3 = \left(\frac{10(10+1)}{2}\right)^2 = \left(\frac{10 \cdot 11}{2}\right)^2 = (5 \cdot 11)^2 = 55^2 = 3025$. $\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$.

Step 4: Calculate the Sum and Match the Target Form. Substitute these values back into the expression for $S$: $S = 11 - 8(3025) + 4(385)$ $S = 11 - 24200 + 1540$ $S = 11 - (24200 - 1540)$ $S = 11 - 22660$

We are given that $S = \alpha - 220\beta$. We need to express $22660$ in the form $220\beta$. Let's divide $22660$ by $220$: $22660 \div 220 = \frac{2266}{22} = \frac{1133}{11} = 103$ So, $22660 = 220 \times 103$.

Substituting this back into the sum: $S = 11 - 220 \times 103$

Comparing this with $S = \alpha - 220\beta$, we can identify: $\alpha = 11$ $\beta = 103$

Thus, the ordered pair $(\alpha, \beta)$ is $(11, 103)$.

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Common Mistakes & Tips

• Incorrectly identifying the general term: Carefully observe the pattern, especially how the numbers change (even/odd, squares, consecutive terms).
• Off-by-one error in summation limits: Ensure the range of $r$ correctly covers all terms in the series.
• Algebraic errors during simplification: Double-check expansions and subtractions, especially with negative signs.
• Calculation errors: Use a calculator for large numbers if allowed, or be meticulous with manual calculations. Alternatively, look for opportunities to factor out common terms early, as shown in the alternative calculation below.

Alternative Calculation for Step 4: From $S = 11 - 8\sum_{r=1}^{10} r^3 + 4\sum_{r=1}^{10} r^2$: $S = 11 - 8(3025) + 4(385)$ We know $3025 = 55^2$ and $385 = 55 \times 7$. $S = 11 - 8(55^2) + 4(55 \times 7)$ We want to factor out $220 = 4 \times 55$. $S = 11 - 2 \cdot (4 \cdot 55) \cdot 55 + (4 \cdot 55) \cdot 7$ $S = 11 - 2 \cdot (220) \cdot 55 + (220) \cdot 7$ $S = 11 - 220 (2 \cdot 55 - 7)$ $S = 11 - 220 (110 - 7)$ $S = 11 - 220 (103)$ This yields $\alpha = 11$ and $\beta = 103$.

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Summary

The problem involved calculating the sum of a series with a specific pattern. We identified the general term of the series as $T_r = 1 - (2r)^2(2r-1)$ for $r=1$ to $10$, with an initial term of $1$. After simplifying the general term to $1 - 8r^3 + 4r^2$, we used standard summation formulas for $r^2$ and $r^3$ with $n=10$. The calculated sum was then expressed in the required form $\alpha - 220\beta$, leading to the values $\alpha = 11$ and $\beta = 103$.

The final answer is $\boxed{\text{(11, 103)}}$, which corresponds to option (A).