Key Concepts and Formulas

This problem relies heavily on the fundamental properties of binomial coefficients derived from the Binomial Theorem. Specifically, we will use:

1. Sum of all binomial coefficients: For any non-negative integer $n$, the sum of all binomial coefficients is given by: $\sum_{r=0}^{n} {^{n}{C_r}} = {^{n}{C_0}} + {^{n}{C_1}} + \dots + {^{n}{C_n}} = 2^n$ This is obtained by setting $x=1$ in the binomial expansion of $(1+x)^n$.

2. Sum of even-indexed or odd-indexed binomial coefficients: For any positive integer $n$, the sum of even-indexed binomial coefficients and the sum of odd-indexed binomial coefficients are both equal to $2^{n-1}$: ${^{n}{C_0}} + {^{n}{C_2}} + {^{n}{C_4}} + \dots = 2^{n-1}$ ${^{n}{C_1}} + {^{n}{C_3}} + {^{n}{C_5}} + \dots = 2^{n-1}$ These can be derived by adding and subtracting the expansions of $(1+1)^n$ and $(1-1)^n$.

3. Difference of Squares Identity: An algebraic identity that states: $(a+b)(a-b) = a^2 - b^2$

Step-by-Step Derivation

Let the given expression be $L$. $L = 1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}})$

Step 1: Simplify the first bracket term Consider the term $(2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})$. We know that ${}^{49}{C_0} = 1$. We can rewrite the term by factoring out one of the '2's and using this identity: $(2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}}) = (1 + 1 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})$ Substitute ${}^{49}{C_0} = 1$: $= (1 + {}^{49}{C_0} + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})$ Now, the expression inside the parenthesis is the sum of all binomial coefficients for $n=49$. Using the first key concept: ${}^{49}{C_0} + {}^{49}{C_1} + \dots + {}^{49}{C_{49}} = 2^{49}$ So, the first bracket simplifies to: $(1 + 2^{49})$

Step 2: Simplify the second bracket term Consider the term ${}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}}$. This is the sum of even-indexed binomial coefficients for $n=50$, but it starts from ${}^{50}{C_2}$ instead of ${}^{50}{C_0}$. From the second key concept, the sum of all even-indexed binomial coefficients for $n=50$ is: ${}^{50}{C_0} + {}^{50}{C_2} + {}^{50}{C_4} + \dots + {}^{50}{C_{50}} = 2^{50-1} = 2^{49}$ To find the value of the given term, we subtract ${}^{50}{C_0}$ from this sum. We know that ${}^{50}{C_0} = 1$. $({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}}) = ({}^{50}{C_0} + {}^{50}{C_2} + \dots + {}^{50}{C_{50}}) - {}^{50}{C_0}$ $= 2^{49} - 1$

Step 3: Combine the simplified terms and calculate L Now substitute the simplified forms of both bracket terms back into the original expression for $L$: $L = 1 + (1 + 2^{49})(2^{49} - 1)$ This expression is in the form $1 + (a+b)(a-b)$, where $a=2^{49}$ and $b=1$. Using the difference of squares identity $(a+b)(a-b) = a^2 - b^2$: $(1 + 2^{49})(2^{49} - 1) = (2^{49})^2 - 1^2$ $= 2^{(49 \times 2)} - 1$ $= 2^{98} - 1$ Substitute this back into the expression for $L$: $L = 1 + (2^{98} - 1)$ $L = 1 + 2^{98} - 1$ $L = 2^{98}$

Step 4: Determine the values of n and m The problem states that $L = 2^n \cdot m$, where $m$ is an odd number. We found $L = 2^{98}$. Comparing this with $2^n \cdot m$: $2^{98} = 2^n \cdot m$ For this equality to hold, we must have $n=98$ and $m=1$. The condition that $m$ must be odd is satisfied, as $1$ is an odd number.

Step 5: Calculate n + m Finally, we need to find the value of $n+m$: $n+m = 98 + 1 = 99$

Tips for Success

• Recognize patterns: Always be on the lookout for standard binomial identities. Terms like $C_0 + C_1 + \dots$ or $C_0 + C_2 + \dots$ are strong indicators to use the sum properties.
• Adjust for missing terms: If a sum of binomial coefficients doesn't start from $C_0$ or includes/excludes specific terms, adjust it by adding or subtracting the known values of those individual terms (e.g., $C_0 = 1$, $C_1 = n$).
• Algebraic Simplification: Don't forget common algebraic identities like the difference of squares, as they can significantly simplify expressions.

Summary

The problem was efficiently solved by recognizing and applying key properties of binomial coefficients. We first simplified the two complex bracket terms using the sum of all binomial coefficients property and the sum of even-indexed binomial coefficients property, respectively. The resulting expression then simplified further using the difference of squares identity, leading to $2^{98}$. By matching this form with $2^n \cdot m$, we found $n=98$ and $m=1$, which allowed us to calculate $n+m=99$. This problem emphasizes the importance of mastering binomial identities for quick and accurate problem-solving.