Key Concepts and Formulas

This problem primarily relies on the properties and identities of binomial coefficients. The core formulas we will use are:

1. Ratio of Consecutive Binomial Coefficients: $\frac{{ }^{n} C_r}{{ }^{n} C_{r-1}} = \frac{n-r+1}{r}$
2. Product Identity: $r \cdot {^n C_r} = n \cdot {^{n-1} C_{r-1}}$
3. Symmetry Property: ${^n C_k} = {^n C_{n-k}}$
4. Sum of Binomial Coefficients: $\sum_{k=0}^{n} {^n C_k} = 2^n$
5. Sum of Product of Index and Binomial Coefficient: $\sum_{k=0}^{n} k \cdot {^n C_k} = n \cdot 2^{n-1}$

We are given the sum: $S = \sum_{r=1}^{30} \frac{r^2\left({ }^{30} C_r\right)^2}{{ }^{30} C_{r-1}}$

Step-by-Step Solution

1. Simplifying the Ratio of Binomial Coefficients

The first step is to simplify the complex term involving the ratio of binomial coefficients. We can rewrite the general term of the sum as: $T_r = r^2 \cdot \frac{{ }^{30} C_r}{{ }^{30} C_{r-1}} \cdot { }^{30} C_r$

Now, we apply the ratio identity $\frac{{ }^{n} C_r}{{ }^{n} C_{r-1}} = \frac{n-r+1}{r}$ with $n=30$: $\frac{{ }^{30} C_r}{{ }^{30} C_{r-1}} = \frac{30-r+1}{r} = \frac{31-r}{r}$

Substitute this into the expression for $T_r$: $T_r = r^2 \cdot \left(\frac{31-r}{r}\right) \cdot { }^{30} C_r$ $T_r = r(31-r) { }^{30} C_r$

Why this step? Simplifying the ratio of binomial coefficients reduces the complexity of the expression, making subsequent algebraic manipulations more straightforward.

2. Applying the Product Identity for Binomial Coefficients

Next, we will simplify the term $r \cdot { }^{30} C_r$ using the identity $k \cdot {^n C_k} = n \cdot {^{n-1} C_{k-1}}$. Here, $n=30$ and $k=r$. $r \cdot { }^{30} C_r = 30 \cdot { }^{29} C_{r-1}$

Substitute this back into our simplified $T_r$: $T_r = (31-r) \left(r \cdot { }^{30} C_r\right)$ $T_r = (31-r) \left(30 \cdot { }^{29} C_{r-1}\right)$ $T_r = 30(31-r) { }^{29} C_{r-1}$

Now, the sum becomes: $S = \sum_{r=1}^{30} 30(31-r) { }^{29} C_{r-1}$

Why this step? This identity allows us to "reduce" the upper index of the binomial coefficient from $30$ to $29$, which is crucial for applying standard summation formulas later. It also removes the explicit 'r' multiplier, making the term simpler.

3. Changing the Index of Summation

To align the summation with standard binomial coefficient sum formulas, let's perform a change of index. Let $k = r-1$. When $r=1$, $k=1-1=0$. When $r=30$, $k=30-1=29$. Also, $r = k+1$, so $31-r = 31-(k+1) = 30-k$.

Substituting these into the sum: $S = \sum_{k=0}^{29} 30(30-k) { }^{29} C_k$ $S = 30 \sum_{k=0}^{29} (30-k) { }^{29} C_k$

Why this step? Changing the index and limits allows us to work with a summation starting from $k=0$, which is the conventional starting point for many binomial summation identities.

4. Splitting the Sum and Applying Summation Identities

Now, we use the symmetry property ${^n C_k} = {^n C_{n-k}}$. For $n=29$, ${^{29} C_k} = {^{29} C_{29-k}}$. Substitute this into the sum: $S = 30 \sum_{k=0}^{29} (30-k) { }^{29} C_{29-k}$

Let $j = 29-k$. When $k=0$, $j=29$. When $k=29$, $j=0$. Also, $k = 29-j$, so $30-k = 30-(29-j) = 1+j$.

Substituting these new values and reversing the order of summation (which doesn't affect the sum): $S = 30 \sum_{j=0}^{29} (1+j) { }^{29} C_j$

Now, split the sum: $S = 30 \left[ \sum_{j=0}^{29} 1 \cdot { }^{29} C_j + \sum_{j=0}^{29} j \cdot { }^{29} C_j \right]$

Why this step? This transformation, using symmetry and a new index, brings the sum into a form where we can directly apply the standard sum identities for binomial coefficients.

5. Evaluating the Sums

We apply the standard sum identities:

• $\sum_{k=0}^{n} {^n C_k} = 2^n$
• $\sum_{k=0}^{n} k \cdot {^n C_k} = n \cdot 2^{n-1}$

For the first part of our sum, with $n=29$: $\sum_{j=0}^{29} { }^{29} C_j = 2^{29}$

For the second part of our sum, with $n=29$: $\sum_{j=0}^{29} j \cdot { }^{29} C_j = 29 \cdot 2^{29-1} = 29 \cdot 2^{28}$

Substitute these values back into the expression for $S$: $S = 30 \left[ 2^{29} + 29 \cdot 2^{28} \right]$

Why this step? These are fundamental identities for binomial sums, allowing us to compute the value directly without iterating through all terms.

6. Final Simplification to Determine $\alpha$

Now, we simplify the expression to match the given format $\alpha \times 2^{29}$: $S = 30 \left[ 2 \cdot 2^{28} + 29 \cdot 2^{28} \right]$ Factor out $2^{28}$: $S = 30 \left[ (2+29) \cdot 2^{28} \right]$ $S = 30 \left[ 31 \cdot 2^{28} \right]$ $S = 930 \cdot 2^{28}$

To express this in the form $\alpha \times 2^{29}$, we rewrite $930$: $S = (465 \cdot 2) \cdot 2^{28}$ $S = 465 \cdot (2 \cdot 2^{28})$ $S = 465 \cdot 2^{29}$

Comparing this with $\alpha \times 2^{29}$, we find that: $\alpha = 465$

Why this step? The final algebraic manipulation is to simplify the expression and present the result in the specific format required by the problem, isolating the value of $\alpha$.

Important Tips and Common Mistakes

• Master Binomial Identities: This problem is a classic example of how a seemingly complex summation can be dramatically simplified by applying the correct binomial identities. Memorize and understand the common ones.
• Index Manipulation: Be very careful when changing the index of summation. Ensure the new limits and the transformed expression correctly represent the original sum. A common mistake is not adjusting all parts of the term (coefficient, index in $C$, limits) simultaneously.
• Factor out Common Terms: When dealing with powers of 2, always try to factor out the smallest power to simplify additions and subtractions.
• Don't Rush the Simplification: Each step of simplification should be deliberate. One misstep can lead to a completely different result.

Summary and Key Takeaway

This problem demonstrates the power of combining several binomial identities to evaluate a complex summation. By systematically applying the ratio property, product identity, symmetry property, and then the summation formulas, the sum simplifies beautifully. The key takeaway is that recognizing and correctly applying these identities is fundamental for solving problems involving binomial coefficients in competitive mathematics. Always look for ways to reduce the complexity of the terms and align them with known summation patterns.