Given the equation:

$\sum_{r=0}^{2n} a_r (x-2)^r = \sum_{r=0}^{2n} b_r (x-3)^r$

and the condition $a_k = 1$ for all $k \geq n$, we are to explore the relationship between the coefficients $a_r$ and $b_r$.

Step 1: Understand the polynomial expressions

The given expressions represent polynomials in terms of $(x-2)$ on the left side and $(x-3)$ on the right side.

Step 2: Expand the polynomial expressions

Let's expand the left-hand side:

$\sum_{r=0}^{2n} a_r (x-2)^r = a_0 + a_1(x-2) + a_2(x-2)^2 + \cdots + a_{2n}(x-2)^{2n}$

Similarly, expand the right-hand side:

$\sum_{r=0}^{2n} b_r (x-3)^r = b_0 + b_1(x-3) + b_2(x-3)^2 + \cdots + b_{2n}(x-3)^{2n}$

Since these two polynomials are equal for all $x$, the coefficients of corresponding powers of $x$ must be equal.

Step 3: Compare coefficients of like powers of $x$

To compare the coefficients, we express $(x-2)^r$ and $(x-3)^r$ in terms of powers of $x$.

Using the binomial expansion:

$(x-2)^r = \sum_{k=0}^{r} \binom{r}{k} (-2)^{r-k} x^k$

$(x-3)^r = \sum_{k=0}^{r} \binom{r}{k} (-3)^{r-k} x^k$

Substituting these into the original expressions, the coefficients of $x^k$ from both sides must be equal:

$\sum_{r=k}^{2n} a_r \binom{r}{k} (-2)^{r-k} = \sum_{r=k}^{2n} b_r \binom{r}{k} (-3)^{r-k}$

Step 4: Apply the condition $a_k = 1$ for $k \geq n$

Given $a_k = 1$ for $k \geq n$, for $r \geq n$, the left-hand sum becomes:

$\sum_{r=k}^{2n} \binom{r}{k} (-2)^{r-k}$

The right-hand side remains:

$\sum_{r=k}^{2n} b_r \binom{r}{k} (-3)^{r-k}$

Step 5: Analyze specific cases (e.g., for $r = n$)

Let’s consider the specific case $r = n$ where $a_n = 1$:

$\sum_{r=n}^{2n} \binom{r}{n} (-2)^{r-n} = \sum_{r=n}^{2n} b_r \binom{r}{n} (-3)^{r-n}$

This simplifies to:

$\sum_{r=n}^{2n} \binom{r}{n} (-2)^{r-n} = b_n (-3)^0 + \text{other terms}$

Since $(-3)^0 = 1$, we have:

$b_n = \sum_{r=n}^{2n} \binom{r}{n} (-2)^{r-n}$

Step 6: Relate $b_n$ to binomial coefficients

The expression for $b_n$ corresponds to a binomial sum, which can be written as:

$b_n = \sum_{r=n}^{2n} \binom{r}{n} (-2)^{r-n} = \binom{2n+1}{n+1}$

This final result shows that $b_n = \binom{2n+1}{n+1}$, which is a combinatorial identity.

Conclusion

Thus, the comparison of the polynomial coefficients shows that $b_n = \binom{2n+1}{n+1}$.

Would you like more details on any specific step or have any other questions?

Here are five related questions:

1. What is the binomial theorem, and how does it apply to polynomial expansions?
2. How do you derive identities involving binomial coefficients?
3. How can generating functions be used to analyze sequences and their sums?
4. What are the properties of binomial coefficients in combinatorics?
5. How does the inclusion-exclusion principle relate to binomial sums?

Tip: When expanding polynomials with binomial terms, carefully compare the coefficients of like powers of $x$ to identify key relationships and patterns.