Key Concepts and Formulas

This problem primarily relies on fundamental identities and properties of binomial coefficients, specifically:

1. Symmetry Property: ${}^n C_r = {}^n C_{n-r}$. This means the binomial coefficient for choosing $r$ items from $n$ is the same as choosing $n-r$ items from $n$.
2. Identity involving $r \cdot {}^n C_r$: $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$. This identity is very useful for simplifying sums where a term is multiplied by its index $r$.
3. Sum of Squares of Binomial Coefficients: $\sum_{r=0}^{n} ({}^n C_r)^2 = {}^{2n} C_n$. This identity states that the sum of the squares of binomial coefficients for a given $n$ is equal to the central binomial coefficient for $2n$.
4. Vandermonde's Identity (Combinatorial Interpretation): $\sum_{k=0}^{r} {}^n C_k \cdot {}^m C_{r-k} = {}^{n+m} C_r$. This identity can be understood as choosing $r$ items from a total of $n+m$ items, where $n$ items are of one type and $m$ items are of another. Alternatively, it represents the coefficient of $x^r$ in the product of two binomial expansions: $(1+x)^n \cdot (1+x)^m = (1+x)^{n+m}$.

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Problem Statement

Given the sum: $S = {({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2}$ If $S = {{\alpha 60!} \over {{{(30!)}^2}}}$, we need to find the value of $\alpha$.

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Method 1: Using Symmetry and Sum of Squares Identity

Step 1: Express the sum in sigma notation and extend for convenience. The given sum can be written as: $S = \sum_{r=1}^{30} r ({}^{30}C_r)^2$ To make use of identities that typically sum from $r=0$, we can observe that the term for $r=0$ would be $0 \cdot ({}^{30}C_0)^2 = 0 \cdot (1)^2 = 0$. Adding this term does not change the sum. So, we can write: $S = \sum_{r=0}^{30} r ({}^{30}C_r)^2 \quad (*)$

Step 2: Utilize the symmetry property of binomial coefficients. The symmetry property states ${}^n C_r = {}^n C_{n-r}$. For $n=30$, we have ${}^{30}C_r = {}^{30}C_{30-r}$. Let's rewrite the sum $S$ by reversing the order of summation and applying this property. If we replace $r$ with $30-r$ in the original sum's index and also use the symmetry property for ${}^{30}C_r$: $S = \sum_{r=0}^{30} (30-r) ({}^{30}C_{30-r})^2$ Since ${}^{30}C_{30-r} = {}^{30}C_r$, we get: $S = \sum_{r=0}^{30} (30-r) ({}^{30}C_r)^2 \quad (**)$ Explanation: We wrote the sum in two ways. The first way iterates $r$ from $0$ to $30$ and multiplies $r$ with $({}^{30}C_r)^2$. The second way iterates $r$ from $0$ to $30$ but effectively considers the terms from $r=30$ down to $r=0$. For example, the term $r=0$ in the second sum corresponds to $30 \cdot ({}^{30}C_0)^2$, which is the coefficient $30$ from the largest index ($30-0$), while the coefficient for $r=30$ is $0 \cdot ({}^{30}C_{30})^2$. This technique is common for sums involving terms multiplied by their indices.

Step 3: Add the two forms of the sum. Now, let's add equations $(*)$ and $(**)$ term by term: $S + S = \sum_{r=0}^{30} r ({}^{30}C_r)^2 + \sum_{r=0}^{30} (30-r) ({}^{30}C_r)^2$ $2S = \sum_{r=0}^{30} [r + (30-r)] ({}^{30}C_r)^2$ $2S = \sum_{r=0}^{30} 30 ({^{30}C_r})^2$ $2S = 30 \sum_{r=0}^{30} ({^{30}C_r})^2$ Explanation: By adding the sums, the terms involving $r$ cancel out in the coefficient, leaving a constant $30$ multiplying each squared binomial coefficient. This simplifies the sum significantly.

Step 4: Apply the sum of squares identity. We know that $\sum_{r=0}^{n} ({}^n C_r)^2 = {}^{2n} C_n$. In our case, $n=30$, so $\sum_{r=0}^{30} ({}^{30}C_r)^2 = {}^{2 \times 30} C_{30} = {}^{60} C_{30}$. Substituting this into our equation for $2S$: $2S = 30 \cdot {}^{60} C_{30}$ $S = 15 \cdot {}^{60} C_{30}$

Step 5: Convert to factorial form and find $\alpha$. Recall that ${}^n C_r = \frac{n!}{r!(n-r)!}$. So, ${}^{60} C_{30} = \frac{60!}{30!(60-30)!} = \frac{60!}{30!30!} = \frac{60!}{(30!)^2}$. Substituting this back into the expression for $S$: $S = 15 \cdot \frac{60!}{(30!)^2}$ We are given that $S = {{\alpha 60!} \over {{{(30!)}^2}}}$. Comparing the two expressions for $S$: $15 \cdot \frac{60!}{(30!)^2} = \frac{\alpha 60!}{(30!)^2}$ By comparing the coefficients, we find: $\alpha = 15$

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Method 2: Using $r \cdot {}^n C_r$ Identity and Vandermonde's Identity

Step 1: Express the sum in sigma notation. The given sum is: $S = \sum_{r=1}^{30} r ({}^{30}C_r)^2$ We can rewrite $({}^{30}C_r)^2$ as ${}^{30}C_r \cdot {}^{30}C_r$: $S = \sum_{r=1}^{30} r \cdot {}^{30}C_r \cdot {}^{30}C_r$

Step 2: Apply the identity $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$. For $n=30$, this identity becomes $r \cdot {}^{30}C_r = 30 \cdot {}^{29}C_{r-1}$. This identity is crucial because it allows us to eliminate the index $r$ from the coefficient term, replacing it with a constant $n$ and modifying the binomial coefficient. Substitute this into the sum: $S = \sum_{r=1}^{30} (30 \cdot {}^{29}C_{r-1}) \cdot {}^{30}C_r$ Factor out the constant $30$: $S = 30 \sum_{r=1}^{30} {}^{29}C_{r-1} \cdot {}^{30}C_r$

Step 3: Prepare for Vandermonde's Identity (Coefficient Extraction). Vandermonde's Identity is of the form $\sum_{k} {}^n C_k \cdot {}^m C_{r-k} = {}^{n+m} C_r$. Our current sum is $\sum_{r=1}^{30} {}^{29}C_{r-1} \cdot {}^{30}C_r$. Let $k = r-1$. When $r=1$, $k=0$. When $r=30$, $k=29$. So the sum becomes: $S = 30 \sum_{k=0}^{29} {}^{29}C_k \cdot {}^{30}C_{k+1}$ Now, we need to manipulate the second binomial coefficient ${}^{30}C_{k+1}$ to fit the form ${}^m C_{P-k}$. We can use the symmetry property ${}^n C_r = {}^n C_{n-r}$ again: ${}^{30}C_{k+1} = {}^{30}C_{30 - (k+1)} = {}^{30}C_{29-k}$. Substitute this back into the sum: $S = 30 \sum_{k=0}^{29} {}^{29}C_k \cdot {}^{30}C_{29-k}$ Explanation: We transformed the sum into a form where the sum of the lower indices of the binomial coefficients is constant ($k + (29-k) = 29$). This is the characteristic structure for applying Vandermonde's Identity. The sum represents the coefficient of $x^{29}$ in the product of two polynomial expansions: $(1+x)^{29}$ and $(1+x)^{30}$.

Step 4: Apply Vandermonde's Identity. Using Vandermonde's Identity, $\sum_{k=0}^{r} {}^n C_k \cdot {}^m C_{r-k} = {}^{n+m} C_r$. In our sum: $n=29$, $m=30$, and $r=29$. Therefore, $\sum_{k=0}^{29} {}^{29}C_k \cdot {}^{30}C_{29-k} = {}^{29+30} C_{29} = {}^{59} C_{29}$. So, $S = 30 \cdot {}^{59} C_{29}$

Step 5: Relate ${}^{59} C_{29}$ to ${}^{60} C_{30}$. We need to express ${}^{59} C_{29}$ in terms of ${}^{60} C_{30}$ to match the required format. We know that ${}^n C_r = \frac{n}{r} {}^{n-1} C_{r-1}$. Let $n=60$ and $r=30$: ${}^{60} C_{30} = \frac{60}{30} {}^{60-1} C_{30-1} = 2 \cdot {}^{59} C_{29}$. From this, we can write ${}^{59} C_{29} = \frac{1}{2} {}^{60} C_{30}$. Substitute this into the expression for $S$: $S = 30 \cdot \left(\frac{1}{2} {}^{60} C_{30}\right)$ $S = 15 \cdot {}^{60} C_{30}$

Step 6: Convert to factorial form and find $\alpha$. As in Method 1, convert ${}^{60} C_{30}$ to factorial form: ${}^{60} C_{30} = \frac{60!}{30!30!} = \frac{60!}{(30!)^2}$. Substitute this into the expression for $S$: $S = 15 \cdot \frac{60!}{(30!)^2}$ Comparing this with the given $S = {{\alpha 60!} \over {{{(30!)}^2}}}$, we find: $\alpha = 15$

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Tips and Common Mistakes

• Forgetting $0 \cdot ({}^n C_0)^2 = 0$: Often, sums start from $r=1$. Adding the $r=0$ term (if it's zero) can make it easier to apply standard identities that sum from $0$ to $n$.
• Incorrectly applying identities: Ensure you understand the conditions and parameters for each binomial identity. For example, Vandermonde's Identity requires the sum of the lower indices to be constant.
• Algebraic errors: Be careful with factorization and simplification of factorials.
• Not recognizing connection between identities: The ability to transform ${}^{n} C_r$ into related forms like ${}^{n} C_{n-r}$ or $\frac{n}{r}{}^{n-1} C_{r-1}$ is key to solving such problems efficiently.

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Summary and Key Takeaway

This problem demonstrates the power and elegance of binomial coefficient identities in simplifying complex sums. Both methods lead to the same result. Method 1 (summation and symmetry) is intuitive for problems with a clear pattern of coefficients. Method 2 (using $r \cdot {}^n C_r$ and Vandermonde's Identity) is a more general and powerful technique, especially when an index $r$ is a factor in the sum. Mastering these identities and their applications is crucial for solving problems in combinatorics and the binomial theorem. The final answer is $\alpha = 15$.