Evaluate: C 0 + C 1 2 + C 2 3 + ⋯ C N N + 1 C_0+\frac{C_1}2+\frac{C_2}3+\cdots\frac{C_n}{n+1} C 0 ​ + 2 C 1 ​ ​ + 3 C 2 ​ ​ + ⋯ N + 1 C N ​ ​

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Introduction

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In this article, we will delve into the evaluation of a summation involving binomial coefficients. The given expression is $C_0+\frac{C_1}2+\frac{C_2}3+\cdots\frac{C_n}{n+1}$, where $C_k$ denotes the usual binomial coefficient. We will explore the solution to this problem and compare it with the given answer.

Understanding Binomial Coefficients

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Before we proceed with the evaluation, let's briefly discuss binomial coefficients. The binomial coefficient $C_k$ is defined as:

$$C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$

where $n!$ denotes the factorial of $n$. The binomial coefficient represents the number of ways to choose $k$ elements from a set of $n$ elements.

Evaluating the Summation

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To evaluate the given summation, we can start by writing out the first few terms:

$$C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \cdots + \frac{C_n}{n+1}$$

We can rewrite each term using the definition of the binomial coefficient:

$$\frac{n!}{0!(n-0)!} + \frac{n!}{1!(n-1)!} \cdot \frac{1}{2} + \frac{n!}{2!(n-2)!} \cdot \frac{1}{3} + \cdots + \frac{n!}{n!(n-n)!} \cdot \frac{1}{n+1}$$

Simplifying each term, we get:

$$n + \frac{n(n-1)}{2} \cdot \frac{1}{2} + \frac{n(n-1)(n-2)}{6} \cdot \frac{1}{3} + \cdots + \frac{n!}{n!} \cdot \frac{1}{n+1}$$

Simplifying the Expression

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We can simplify the expression further by canceling out common factors:

$$n + \frac{n(n-1)}{4} + \frac{n(n-1)(n-2)}{24} + \cdots + \frac{1}{n+1}$$

Using the Binomial Theorem

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We can use the binomial theorem to rewrite the expression:

$$(1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k}$$

Expanding the binomial, we get:

$$2^n = \sum_{k=0}^{n} \binom{n}{k}$$

Relating to the Original Expression

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We can relate the binomial expansion to the original expression by multiplying both sides by $\frac{1}{n+1}$:

$$\frac{2^n}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{1}{n+1}$$

Comparing with the Given Answer

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The given answer is $\frac{2^{n+1}}{n+1}$. We can see that our solution is close, but not exactly the same. Let's examine the difference:

$$frac{2^{n+1}}{n+1} - \frac{2^n}{n+1} = \frac{2^n}{n+1}$$

Conclusion

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In conclusion, we have evaluated the summation of binomial coefficients and compared it with the given answer. Our solution is $\frac{2^n}{n+1}$, while the given answer is $\frac{2^{n+1}}{n+1}$. The difference between the two solutions is $\frac{2^n}{n+1}$.

Final Answer

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The final answer is $\boxed{\frac{2^n}{n+1}}$.

Additional Information

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The given answer can be obtained by using the binomial theorem and multiplying both sides by $\frac{1}{n+1}$. This results in:

$$\frac{2^{n+1}}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{1}{n+1}$$

This is the same as the original expression, but with an additional factor of $2$ in the numerator.

Comparison with the Given Answer

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We can compare our solution with the given answer by examining the difference:

$$\frac{2^{n+1}}{n+1} - \frac{2^n}{n+1} = \frac{2^n}{n+1}$$

This shows that our solution is close to the given answer, but not exactly the same.

Final Thoughts

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In conclusion, we have evaluated the summation of binomial coefficients and compared it with the given answer. Our solution is $\frac{2^n}{n+1}$, while the given answer is $\frac{2^{n+1}}{n+1}$. The difference between the two solutions is $\frac{2^n}{n+1}$.

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Introduction

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In our previous article, we evaluated the summation of binomial coefficients and compared it with the given answer. In this article, we will provide a Q&A section to address any questions or concerns that readers may have.

Q: What is the definition of a binomial coefficient?

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A: A binomial coefficient, denoted as $C_k$, is defined as:

$$C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$

where $n!$ denotes the factorial of $n$.

Q: How do I evaluate the summation of binomial coefficients?

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A: To evaluate the summation of binomial coefficients, you can use the definition of the binomial coefficient and simplify the expression. You can also use the binomial theorem to rewrite the expression.

Q: What is the binomial theorem?

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A: The binomial theorem is a mathematical formula that describes the expansion of a binomial raised to a power. It is given by:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Q: How do I use the binomial theorem to evaluate the summation of binomial coefficients?

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A: To use the binomial theorem to evaluate the summation of binomial coefficients, you can multiply both sides of the equation by $\frac{1}{n+1}$:

$$\frac{(1 + 1)^n}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{1}{n+1}$$

Q: What is the final answer to the summation of binomial coefficients?

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A: The final answer to the summation of binomial coefficients is $\boxed{\frac{2^n}{n+1}}$.

Q: Why is the given answer $\frac{2^{n+1}}{n+1}$?

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A: The given answer $\frac{2^{n+1}}{n+1}$ can be obtained by using the binomial theorem and multiplying both sides by $\frac{1}{n+1}$. This results in:

$$\frac{2^{n+1}}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{1}{n+1}$$

Q: What is the difference between the two solutions?

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A: The difference between the two solutions is $\frac{2^n}{n+1}$.

Q: Can you provide a step-by-step solution to the problem?

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A: Yes, here is a step-by-step solution to the problem:

1. Evaluate the binomial coefficient $C_k$ using the definition:

$$C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$

2. Simplify the expression by canceling out common factors:

$$n + \frac{n(n-1)}{4} + \frac{n(n-1)(n-2)}{24} + \cdots + \frac{1}{n+1}$$

3. Use the binomial theorem to rewrite the expression:

$$(1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k}$$

4. Multiply both sides of the equation by $\frac{1}{n+1}$:

$$\frac{2^n}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{1}{n+1}$$

Conclusion

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In conclusion, we have provided a Q&A section to address any questions or concerns that readers may have regarding the evaluation of the summation of binomial coefficients. We hope that this article has been helpful in clarifying any doubts and providing a clear understanding of the solution.