Key Concept: The Integral Property of Binomial Coefficients

This problem leverages a fundamental identity related to binomial coefficients involving an integral form. The key property states that for any non-negative integers $n$ and $r$: $\frac{1}{r+1} {}^{n} \mathrm{C}_{r} = \frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1}$ Let's quickly understand why this identity holds true, as it's crucial for solving the problem. We know the definition of a binomial coefficient is ${}^{n} \mathrm{C}_{r} = \frac{n!}{r!(n-r)!}$. Consider the left-hand side: $\frac{1}{r+1} {}^{n} \mathrm{C}_{r} = \frac{1}{r+1} \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)r!(n-r)!} = \frac{n!}{(r+1)!(n-r)!}$ Now, consider the right-hand side: $\frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!((n+1)-(r+1))!} = \frac{1}{n+1} \cdot \frac{(n+1)n!}{(r+1)!(n-r)!}$ $= \frac{n!}{(r+1)!(n-r)!}$ Since both sides simplify to the same expression, the identity is proven. This identity allows us to transform a sum involving $\frac{1}{r+1} {}^{n} \mathrm{C}_{r}$ into a sum of standard binomial coefficients, which we can then evaluate using the binomial theorem.

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Problem Statement and Initial Setup

We are given the equation: $\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$ To make this sum easier to work with, we first rewrite it using summation notation. It's important to recognize the pattern of the terms. If we consider the general term in reverse order (as in the given solution), starting from ${}^{n} \mathrm{C}_{0}$, we have $\frac{1}{0+1}{}^{n}\mathrm{C}_{0}$, then $\frac{1}{1+1}{}^{n}\mathrm{C}_{1}$, and so on, up to $\frac{1}{n+1}{}^{n}\mathrm{C}_{n}$. So, the sum can be compactly written as: $\sum_{r=0}^{n} \frac{1}{r+1} {}^{n} \mathrm{C}_{r} = \frac{1023}{10}$ This step simplifies the representation and prepares the expression for applying the key identity.

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Applying the Integral Property

Now, we apply the integral property of binomial coefficients that we discussed earlier: $\frac{1}{r+1} {}^{n} \mathrm{C}_{r} = \frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1}$. We substitute this into our summation: $\sum_{r=0}^{n} \left( \frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1} \right) = \frac{1023}{10}$ The term $\frac{1}{n+1}$ is a constant with respect to $r$, so we can factor it out of the summation: $\frac{1}{n+1} \sum_{r=0}^{n} {}^{n+1} \mathrm{C}_{r+1} = \frac{1023}{10}$ This transformation is crucial because it converts the sum into a form that is directly related to the sum of binomial coefficients of power $(n+1)$.

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Summing the Binomial Coefficients

Let's expand the summation $\sum_{r=0}^{n} {}^{n+1} \mathrm{C}_{r+1}$:

• When $r=0$, the term is ${}^{n+1} \mathrm{C}_{0+1} = {}^{n+1} \mathrm{C}_{1}$.
• When $r=1$, the term is ${}^{n+1} \mathrm{C}_{1+1} = {}^{n+1} \mathrm{C}_{2}$.
• ...
• When $r=n$, the term is ${}^{n+1} \mathrm{C}_{n+1}$. So the sum becomes: ${}^{n+1} \mathrm{C}_{1} + {}^{n+1} \mathrm{C}_{2} + \ldots + {}^{n+1} \mathrm{C}_{n+1}$ We know from the Binomial Theorem that the sum of all binomial coefficients for a power $k$ is $2^k$. That is: ${}^{k} \mathrm{C}_{0} + {}^{k} \mathrm{C}_{1} + {}^{k} \mathrm{C}_{2} + \ldots + {}^{k} \mathrm{C}_{k} = 2^k$ In our case, $k = n+1$. So, the sum of all binomial coefficients for power $(n+1)$ is $2^{n+1}$: ${}^{n+1} \mathrm{C}_{0} + {}^{n+1} \mathrm{C}_{1} + {}^{n+1} \mathrm{C}_{2} + \ldots + {}^{n+1} \mathrm{C}_{n+1} = 2^{n+1}$ The sum we are interested in is missing the first term, ${}^{n+1} \mathrm{C}_{0}$. Since ${}^{n+1} \mathrm{C}_{0} = 1$ (any number choose 0 is 1), we can write our sum as: $\left( {}^{n+1} \mathrm{C}_{0} + {}^{n+1} \mathrm{C}_{1} + \ldots + {}^{n+1} \mathrm{C}_{n+1} \right) - {}^{n+1} \mathrm{C}_{0} = 2^{n+1} - 1$ Substituting this back into our equation: $\frac{1}{n+1} (2^{n+1} - 1) = \frac{1023}{10}$

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Solving for n

Now we have a straightforward equation to solve for $n$: $\frac{2^{n+1}-1}{n+1} = \frac{1023}{10}$ To find $n$, we need to express $1023$ in the form $2^k - 1$. Let's add 1 to 1023: $1023 + 1 = 1024$. We recognize that $1024$ is a power of 2. Specifically, $2^{10} = 1024$. So, $1023 = 2^{10} - 1$. Substituting this back into the equation: $\frac{2^{n+1}-1}{n+1} = \frac{2^{10}-1}{10}$ By comparing the structure of both sides, we can infer that the exponent and the denominator must be equal: $n+1 = 10$ Solving for $n$: $n = 10 - 1$ $n = 9$

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Tips and Common Mistakes

• Understanding the Identity: The identity $\frac{1}{r+1} {}^{n} \mathrm{C}_{r} = \frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1}$ is not always immediately obvious. If you forget it, you can always derive it using the factorial definition of binomial coefficients, as shown in the "Key Concept" section.
• Summation Limits: Pay close attention to the starting and ending values of $r$ in the summation. When transforming $\sum_{r=0}^{n} {}^{n+1} \mathrm{C}_{r+1}$, ensure you correctly identify which terms are included and which are missing from the full binomial expansion sum $2^{n+1}$. Forgetting to subtract ${}^{n+1} \mathrm{C}_{0}$ is a common error.
• Recognizing Powers of 2: Knowing common powers of 2 (like $2^9=512$, $2^{10}=1024$) can significantly speed up the final step of solving for $n$.
• Generalization: This problem is a specific application of a more general identity for sums of binomial coefficients involving division by $r+1$. This pattern often appears in competitive exams.

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Summary and Key Takeaway

This problem beautifully illustrates how a seemingly complex sum of binomial coefficients can be simplified using a clever identity. By transforming $\frac{1}{r+1} {}^{n} \mathrm{C}_{r}$ into $\frac{1}{n+1} {}^{n+1} \mathrm{C}_{r+1}$, we convert the sum into a standard binomial expansion (minus one term) that can be easily evaluated as $2^{n+1}-1$. The final step involves recognizing powers of 2 to solve for $n$. The key takeaway is to be familiar with integral identities of binomial coefficients and how they relate to the sum of binomial coefficients, $2^n$.

The final answer is $\boxed{\text{9}}$.