Introduction to Key Concepts

This problem requires us to evaluate a sum involving binomial coefficients. We will primarily use the following key properties and techniques:

1. Binomial Theorem: The expansion of $(a+b)^n = \sum_{r=0}^{n} {}^{n}{C_r} a^{n-r} b^r$.
2. Identity for $r \cdot {}^{n}{C_r}$: For $r \ge 1$, we have $r \cdot {}^{n}{C_r} = n \cdot {}^{n-1}{C_{r-1}}$. This identity is crucial for simplifying sums where a coefficient is multiplied by its index.
3. Alternating Sum of Binomial Coefficients: For $n \ge 1$, $\sum_{r=0}^{n} (-1)^r {}^{n}{C_r} = {}^{n}{C_0} - {}^{n}{C_1} + {}^{n}{C_2} - \dots + (-1)^n {}^{n}{C_n} = 0$. This property arises directly from setting $x=1$ in the expansion of $(1-x)^n$.
4. Sum of Odd/Even Binomial Coefficients: For $n \ge 1$, the sum of odd-indexed binomial coefficients equals the sum of even-indexed binomial coefficients, and both are equal to $2^{n-1}$: ${}^{n}{C_1} + {}^{n}{C_3} + {}^{n}{C_5} + \dots = 2^{n-1}$ ${}^{n}{C_0} + {}^{n}{C_2} + {}^{n}{C_4} + \dots = 2^{n-1}$
5. Symmetry Property: ${}^{n}{C_r} = {}^{n}{C_{n-r}}$.

The problem consists of two distinct parts that need to be evaluated and then summed.

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Part 1: Evaluating the Alternating Series with $r \cdot {}^{n}{C_r}$

We need to evaluate the series: $S_1 = - {}^{15}{C_1} + 2.{}^{15}{C_2} - 3.{}^{15}{C_3} + \dots - 15.{}^{15}{C_{15}}$

Method 1: Using the Identity $r \cdot {}^{n}{C_r} = n \cdot {}^{n-1}{C_{r-1}}$

Step 1: Express the series in summation notation. The given series can be written as: $S_1 = \sum_{r=1}^{15} (-1)^r \cdot r \cdot {}^{15}{C_r}$

• Explanation: The term $r \cdot {}^{15}{C_r}$ is present, and the signs alternate, starting with negative for $r=1$. Thus, $(-1)^r$ correctly captures the alternating sign pattern.

Step 2: Apply the identity $r \cdot {}^{n}{C_r} = n \cdot {}^{n-1}{C_{r-1}}$. Here, $n=15$, so we replace $r \cdot {}^{15}{C_r}$ with $15 \cdot {}^{14}{C_{r-1}}$. $S_1 = \sum_{r=1}^{15} (-1)^r \cdot 15 \cdot {}^{14}{C_{r-1}}$

• Explanation: This identity is fundamental for simplifying sums where the index $r$ is multiplied by the binomial coefficient ${}^{n}{C_r}$. It reduces the upper index of the binomial coefficient from $n$ to $n-1$, which often simplifies the overall sum.

Step 3: Factor out the constant and adjust the summation index. $S_1 = 15 \sum_{r=1}^{15} (-1)^r {}^{14}{C_{r-1}}$ Let $k = r-1$. When $r=1$, $k=0$. When $r=15$, $k=14$. Also, $r = k+1$. $S_1 = 15 \sum_{k=0}^{14} (-1)^{k+1} {}^{14}{C_k}$ $S_1 = 15 \sum_{k=0}^{14} (-1) \cdot (-1)^k {}^{14}{C_k}$ $S_1 = -15 \sum_{k=0}^{14} (-1)^k {}^{14}{C_k}$

• Explanation: Factoring out the constant $15$ is a standard algebraic step. Changing the summation index from $r$ to $k=r-1$ makes the binomial coefficient ${}^{14}{C_k}$ align with the standard form of the alternating sum, where the index starts from 0.

Step 4: Apply the property of alternating sum of binomial coefficients. We know that for any positive integer $n$, $\sum_{k=0}^{n} (-1)^k {}^{n}{C_k} = 0$. Here, $n=14$, so $\sum_{k=0}^{14} (-1)^k {}^{14}{C_k} = {}^{14}{C_0} - {}^{14}{C_1} + {}^{14}{C_2} - \dots + {}^{14}{C_{14}} = 0$. $S_1 = -15 \cdot (0)$ $S_1 = 0$

• Explanation: This is a direct application of the property that the alternating sum of binomial coefficients for any power $n \ge 1$ is zero. It stems from the binomial expansion of $(1-1)^n$.

Method 2: Using Differentiation

Step 1: Consider the binomial expansion of $(1-x)^{15}$. We choose $(1-x)^{15}$ because differentiating it with respect to $x$ will bring down the power of $x$ and introduce alternating signs, matching the structure of $S_1$. $(1-x)^{15} = {}^{15}{C_0} - {}^{15}{C_1}x + {}^{15}{C_2}x^2 - {}^{15}{C_3}x^3 + \dots + (-1)^{15} {}^{15}{C_{15}}x^{15}$

• Explanation: This is the standard binomial expansion where $a=1$ and $b=-x$.

Step 2: Differentiate both sides with respect to $x$. $\frac{d}{dx} (1-x)^{15} = \frac{d}{dx} \left( {}^{15}{C_0} - {}^{15}{C_1}x + {}^{15}{C_2}x^2 - {}^{15}{C_3}x^3 + \dots - {}^{15}{C_{15}}x^{15} \right)$ Applying the chain rule on the left side and term-by-term differentiation on the right side: $15(1-x)^{14}(-1) = -{}^{15}{C_1} + 2{}^{15}{C_2}x - 3{}^{15}{C_3}x^2 + \dots - 15{}^{15}{C_{15}}x^{14}$

• Explanation: Differentiation transforms $x^r$ into $r \cdot x^{r-1}$, creating the $r \cdot {}^{n}{C_r}$ pattern that we need. The derivative of $(1-x)$ is $-1$, which introduces the desired negative sign on the left.

Step 3: Substitute $x=1$ into the differentiated equation. Substituting $x=1$ will make the left side zero and simplify the right side to exactly the series we want to evaluate. $-15(1-1)^{14} = -{}^{15}{C_1} + 2{}^{15}{C_2}(1) - 3{}^{15}{C_3}(1)^2 + \dots - 15{}^{15}{C_{15}}(1)^{14}$ $-15(0)^{14} = -{}^{15}{C_1} + 2{}^{15}{C_2} - 3{}^{15}{C_3} + \dots - 15{}^{15}{C_{15}}$ $0 = -{}^{15}{C_1} + 2{}^{15}{C_2} - 3{}^{15}{C_3} + \dots - 15{}^{15}{C_{15}}$

• Explanation: Setting $x=1$ effectively eliminates all powers of $x$ on the right side, leaving only the coefficients multiplied by their original indices, which perfectly matches $S_1$. The term $(1-1)^{14}$ on the left becomes $0^{14}=0$ (since $14 \ge 1$).

Both methods confirm that $S_1 = 0$.

Tip: For sums involving $r \cdot {}^{n}{C_r}$, the differentiation method is often more direct and less prone to sign errors than using the identity, especially for complex alternating series.

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Part 2: Evaluating the Sum of Odd Binomial Coefficients

We need to evaluate the series: $S_2 = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} + \dots + {}^{14}{C_{11}}$

Step 1: Recall the property for the sum of odd-indexed binomial coefficients. For $n \ge 1$, the sum of all odd-indexed binomial coefficients ${}^{n}{C_r}$ is $2^{n-1}$. For $n=14$, the sum of all odd-indexed binomial coefficients is: ${}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} + \dots + {}^{14}{C_{13}} = 2^{14-1} = 2^{13}$

• Explanation: This property is derived from adding the binomial expansions of $(1+x)^n$ and $(1-x)^n$ and then setting $x=1$. The highest odd index for ${}^{14}{C_r}$ is $r=13$, as ${}^{14}{C_{15}}$, etc., are zero.

Step 2: Compare the required sum $S_2$ with the full sum of odd coefficients. The required sum $S_2$ is: $S_2 = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} + \dots + {}^{14}{C_{11}}$ We can see that $S_2$ is missing the last term from the full sum of odd coefficients, which is ${}^{14}{C_{13}}$. So, we can write: $({}^{14}{C_1} + {}^{14}{C_3} + \dots + {}^{14}{C_{11}}) + {}^{14}{C_{13}} = 2^{13}$ $S_2 + {}^{14}{C_{13}} = 2^{13}$

• Explanation: By recognizing that our target sum is a partial sum of the known full sum, we can express the target sum by subtracting the missing terms.

Step 3: Calculate the missing term ${}^{14}{C_{13}}$. Using the symmetry property ${}^{n}{C_r} = {}^{n}{C_{n-r}}$: ${}^{14}{C_{13}} = {}^{14}{C_{14-13}} = {}^{14}{C_1}$ And ${}^{14}{C_1} = \frac{14!}{1!(14-1)!} = \frac{14!}{1!13!} = 14$. So, ${}^{14}{C_{13}} = 14$.

• Explanation: The symmetry property allows for quick calculation of binomial coefficients like ${}^{n}{C_{n-1}}$, which is simply $n$. This avoids calculating large factorials.

Step 4: Substitute the value of ${}^{14}{C_{13}}$ back into the equation and solve for $S_2$. $S_2 + 14 = 2^{13}$ $S_2 = 2^{13} - 14$

• Explanation: This final step isolates $S_2$, giving us the value of the second part of the problem.

Common Mistake: A common error is to assume the sum of odd terms automatically ends at the highest odd index shown (e.g., ${}^{14}{C_{11}}$). Always verify the range of indices against the general formula for the full sum of odd/even binomial coefficients.

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Combining the Results

The problem asks for the sum of the two parts: $S_1 + S_2$. Total Value $= S_1 + S_2$ Total Value $= 0 + (2^{13} - 14)$ Total Value $= 2^{13} - 14$

Comparing this with the given options, the correct answer is (D).

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Summary and Key Takeaway

The value of the expression is $2^{13} - 14$. This problem effectively tests your understanding of two important categories of binomial series sums:

1. Sums involving $r \cdot {}^{n}{C_r}$ with alternating signs: These can often be solved efficiently using either the identity $r \cdot {}^{n}{C_r} = n \cdot {}^{n-1}{C_{r-1}}$ or by differentiating a suitable binomial expansion and then substituting a specific value for $x$.
2. Sums of odd/even binomial coefficients: These rely on the property that for $n \ge 1$, $\sum_{r \text{ odd}} {}^{n}{C_r} = \sum_{r \text{ even}} {}^{n}{C_r} = 2^{n-1}$. It's important to correctly identify the full range of coefficients in the sum and account for any missing terms.

Mastering these identities and techniques is crucial for solving advanced problems in the Binomial Theorem.