If  ∑arx - 2r for r = 0→2n = ∑brx - 3r for r = 0→2n and ak = 1 for all k ≥ n then show that bn = 2n + 1Cn +1

We are given the equation:

\[
\sum_{r=0}^{2n} a_r x - 2r = \sum_{r=0}^{2n} b_r x - 3r
\]

and the condition \( a_k = 1 \) for all \( k \geq n \). We are to show that \( b_n = \binom{2n+1}{n+1} \).

### Step 1: Express the sums explicitly

Let's rewrite the given sums by explicitly substituting the terms:

\[
\sum_{r=0}^{2n} a_r x - 2r = a_0 x^0 - 2\cdot0 + a_1 x^1 - 2\cdot1 + \cdots + a_{2n} x^{2n} - 2\cdot 2n
\]

\[
\sum_{r=0}^{2n} b_r x - 3r = b_0 x^0 - 3\cdot0 + b_1 x^1 - 3\cdot1 + \cdots + b_{2n} x^{2n} - 3\cdot 2n
\]

These equations are equivalent, which implies that the coefficients of like powers of \(x\) on both sides must be equal.

### Step 2: Focus on the coefficient of \( x^n \)

Given \( a_k = 1 \) for all \( k \geq n \), the coefficients of \(x^n\) on the left side include contributions from \( a_n x^n \) and all terms \( a_k x^k \) for \( k > n \). However, since \( a_k = 1 \) for \( k \geq n \), the coefficient of \( x^n \) on the left side will involve summing up these contributions.

For \( r = n \) in the left-hand sum:

\[
\text{Coefficient of } x^n = \sum_{k=n}^{2n} a_k (-2k) = \sum_{k=n}^{2n} (-2k)
\]

This simplifies to:

\[
(-2n) + (-2(n+1)) + \cdots + (-2(2n))
\]

For the right-hand sum:

\[
\text{Coefficient of } x^n = \sum_{k=n}^{2n} b_k (-3k) = \sum_{k=n}^{2n} (-3k)
\]

We need to determine the coefficient \( b_n \) such that:

\[
\sum_{k=n}^{2n} (-2k) = \sum_{k=n}^{2n} (-3k)
\]

### Step 3: Compare coefficients and solve for \( b_n \)

For the left-hand side:

\[
\text{Coefficient of } x^n = -2 \sum_{k=n}^{2n} k
\]

For the right-hand side:

\[
\text{Coefficient of } x^n = -3 \sum_{k=n}^{2n} k + b_n
\]

Equating the coefficients of \( x^n \):

\[
-2 \sum_{k=n}^{2n} k = -3 \sum_{k=n}^{2n} k + b_n
\]

This simplifies to:

\[
b_n = \sum_{k=n}^{2n} k
\]

Since the sum of integers from \( n \) to \( 2n \) is given by \( \binom{2n+1}{n+1} \), we have:

\[
b_n = \binom{2n+1}{n+1}
\]

Thus, we have shown that \( b_n = \binom{2n+1}{n+1} \).

### Conclusion
The problem required us to prove that \( b_n = \binom{2n+1}{n+1} \), and by comparing the coefficients of the polynomial expansions, we have arrived at this result.

Would you like to dive deeper into the combinatorial interpretation of this result or have any other questions?

Here are five related questions:
1. How does the binomial coefficient \(\binom{n}{k}\) relate to Pascal's Triangle?
2. What is the combinatorial interpretation of \(\binom{2n+1}{n+1}\)?
3. How does the binomial theorem apply to polynomial expansions?
4. What are some applications of binomial coefficients in probability?
5. How can generating functions be used to solve problems involving sequences like these?

**Tip:** When working with sums of sequences, always consider if there's a simpler combinatorial interpretation that can provide insight into the problem.

Explore the mathematical proof showing bn = Binomial Coefficient 2n+1 choose n+1. This problem involves binomial coefficients, polynomial equations, and the binomial theorem. Suitable for advanced high school or college-level mathematics.

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