Understanding the Problem We are asked to evaluate a specific sum involving binomial coefficients, express the result in the form $n \cdot 2^m$ where $n, m$ are natural numbers and $n$ is odd (since $\gcd(2, n) = 1$), and then find the value of $n+m$. The given sum is: $S = 30\left( {\matrix{ {30} \cr 0 \cr } } \right) + 29\left( {\matrix{ {30} \cr 1 \cr } } \right) + ...... + 2\left( {\matrix{ {30} \cr {28} \cr } } \right) + 1\left( {\matrix{ {30} \cr {29} \cr } } \right)$ Here, $\left( {\matrix{ n \cr k \cr } } \right)$ is standard notation for the binomial coefficient ${}^n C_k$.

Key Binomial Identities This problem requires the application of a few fundamental binomial identities:

1. Symmetry Property: ${}^n C_k = {}^n C_{n-k}$
   • Explanation: This identity states that choosing $k$ elements from a set of $n$ elements is the same as choosing $n-k$ elements to exclude from the set. It's often used to simplify sums by transforming the binomial coefficients.
2. Product Identity: $k \cdot {}^n C_k = n \cdot {}^{n-1} C_{k-1}$
   • Explanation: This identity is crucial for simplifying sums where a coefficient $k$ is multiplied by ${}^n C_k$. It can be easily proven by expanding the factorial definitions: $k \cdot \frac{n!}{k!(n-k)!} = \frac{n!}{(k-1)!(n-k)!} = n \cdot \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = n \cdot {}^{n-1} C_{k-1}$
3. Sum of Binomial Coefficients: $\sum_{k=0}^n {}^n C_k = {}^n C_0 + {}^n C_1 + \dots + {}^n C_n = 2^n$
   • Explanation: This is a direct consequence of the Binomial Theorem, obtained by setting $x=1$ in the expansion of $(1+x)^n$. It represents the total number of subsets of a set with $n$ elements.

Step-by-Step Solution

Step 1: Rewriting the Sum using Symmetry The given sum has coefficients that decrease from 30 to 1, while the lower index of the binomial coefficient increases from 0 to 29. $S = 30({}^{30}{C_0}) + 29({}^{30}{C_1}) + .... + 2({}^{30}{C_{28}}) + 1({}^{30}{C_{29}})$ To make the structure more consistent for using the product identity, we apply the symmetry property ${}^n C_k = {}^n C_{n-k}$ with $n=30$. Let's rewrite each term:

• $30({}^{30}{C_0}) = 30({}^{30}{C_{30-0}}) = 30({}^{30}{C_{30}})$
• $29({}^{30}{C_1}) = 29({}^{30}{C_{30-1}}) = 29({}^{30}{C_{29}})$
• ...
• $2({}^{30}{C_{28}}) = 2({}^{30}{C_{30-28}}) = 2({}^{30}{C_2})$
• $1({}^{30}{C_{29}}) = 1({}^{30}{C_{30-29}}) = 1({}^{30}{C_1})$

Substituting these transformed terms back into the sum, we get: $S = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$ Explanation: By using the symmetry property, we've transformed the sum into a form where each term is $r \cdot {}^{30}C_r$ (if we reverse the order and consider $r$ from 1 to 30). This rearranged sum is easier to express in sigma notation and prepare for the next identity.

Now, we can write this sum compactly in sigma notation. Note that the term $0 \cdot {}^{30}C_0$ would be zero, so including or excluding it does not change the sum's value. We can start our sum from $r=1$: $S = \sum_{r=1}^{30} r \cdot {}^{30}C_r$ Explanation: This is a standard way to represent the sum, where $r$ is the index that varies from 1 to 30, corresponding to the coefficient multiplying the binomial term and the lower index of the binomial coefficient.

Step 2: Applying the Identity $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$ Now we apply the product identity $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$. In our sum, $n=30$. For each term $r \cdot {}^{30}C_r$, we replace it with $30 \cdot {}^{29}C_{r-1}$. Substituting this into our sum: $S = \sum_{r=1}^{30} 30 \cdot {}^{29}C_{r-1}$ Explanation: This step is taken to simplify the expression. The product identity is designed to eliminate the variable factor $r$ from outside the binomial coefficient, transforming the sum into a simpler form that can be evaluated using the sum of binomial coefficients property.

Step 3: Summing the Binomial Coefficients First, we can factor out the constant $30$ from the summation: $S = 30 \sum_{r=1}^{30} {}^{29}C_{r-1}$ Let's examine the terms inside the summation by letting $k = r-1$. As $r$ goes from $1$ to $30$, $k$ goes from $0$ to $29$:

• When $r=1$, $k=0$, the term is ${}^{29}C_0$
• When $r=2$, $k=1$, the term is ${}^{29}C_1$
• ...
• When $r=30$, $k=29$, the term is ${}^{29}C_{29}$

So the sum becomes: $S = 30 ({}^{29}C_0 + {}^{29}C_1 + {}^{29}C_2 + \dots + {}^{29}C_{29})$ The expression in the parenthesis is the sum of all binomial coefficients for $n=29$. Using the identity $\sum_{k=0}^n {}^n C_k = 2^n$: ${}^{29}C_0 + {}^{29}C_1 + {}^{29}C_2 + \dots + {}^{29}C_{29} = 2^{29}$ Therefore, the sum $S$ is: $S = 30 \cdot 2^{29}$ Explanation: By factoring out the constant and clearly writing out the terms of the summation, we can identify it as a direct application of the sum of binomial coefficients identity, which greatly simplifies the expression.

Step 4: Expressing in the form $n \cdot 2^m$ The problem requires us to express $S$ in the form $n \cdot 2^m$, where $n, m \in \mathbb{N}$ and $\gcd(2, n) = 1$. The condition $\gcd(2, n) = 1$ means that $n$ must be an odd integer. We have $S = 30 \cdot 2^{29}$. To ensure $n$ is odd, we need to factor out all powers of 2 from 30: $30 = 2 \cdot 15$. Now substitute this back into the expression for $S$: $S = (2 \cdot 15) \cdot 2^{29}$ $S = 15 \cdot (2^1 \cdot 2^{29})$ Using the exponent rule $a^x \cdot a^y = a^{x+y}$: $S = 15 \cdot 2^{1+29}$ $S = 15 \cdot 2^{30}$ Comparing this with the given form $n \cdot 2^m$: We identify $n = 15$ and $m = 30$. Let's verify the conditions:

• $n=15$ is a natural number.
• $m=30$ is a natural number.
• $\gcd(2, n) = \gcd(2, 15) = 1$, as 15 is an odd number. All conditions are satisfied. Explanation: This step is about matching the derived form with the problem's specified format. The crucial detail is to ensure that the $n$ component of $n \cdot 2^m$ is odd, which means factoring out any even components from the numerical coefficient into the power of 2.

Step 5: Calculating $n+m$ Finally, we calculate the value of $n+m$: $n+m = 15 + 30 = 45$ Explanation: This is the final arithmetic step to arrive at the answer requested by the problem statement.

Important Tips and Common Mistakes

• Don't Forget Symmetry: When the coefficients of binomial terms in a sum follow a decreasing or increasing arithmetic progression, the symmetry property ${}^n C_k = {}^n C_{n-k}$ is often the first step to simplify the sum's structure.
• Master the Product Identity: The identity $k \cdot {}^n C_k = n \cdot {}^{n-1} C_{k-1}$ is extremely versatile. Make sure you can apply it correctly, especially with respect to the changed indices.
• Condition $\gcd(2,n)=1$: This means $n$ must be odd. If your initial calculation for $n$ is even, you must factor out all powers of 2 and combine them with the $2^m$ term. Forgetting this can lead to an incorrect value for $n$ and $m$.
• Index Alignment: When applying identities or writing sums in sigma notation, always double-check the starting and ending indices to ensure all terms are correctly included.

Summary and Key Takeaway This problem serves as an excellent illustration of how to systematically solve binomial coefficient sums by leveraging fundamental identities. The process involves:

1. Using the symmetry property to standardize the sum's terms.
2. Applying the product identity $k \cdot {}^n C_k = n \cdot {}^{n-1} C_{k-1}$ to simplify the individual terms.
3. Recognizing and applying the sum of all binomial coefficients $\sum_{k=0}^n {}^n C_k = 2^n$.
4. Carefully manipulating the final expression to match the required format ($n \cdot 2^m$) and adhering to conditions like $n$ being odd.

This multi-step approach transforms a seemingly complex sum into a straightforward calculation, emphasizing the power of binomial identities in combinatorics.