Question

If $\sum_{r=0}^{2n} a_r (x-2)^r = \sum_{r=0}^{2n} b_r (x-3)^r$ and $a_k = 1 \ \forall k \ge n$, show that $b_n = ^{2n+1}C_{n+1}$.

Answer 1

^{2n+1}C_{n+1}

Answer 2

The problem asks us to show that $b_n = ^{2n+1}C_{n+1}$, given two series expansions of a polynomial $P(x)$ and a condition on the coefficients $a_k$.

Let the given polynomial be $P(x)$. We have two representations for $P(x)$:

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

We are given that $a_k = 1$ for all $k \ge n$. We need to find the coefficient $b_n$.

From the second representation, $P(x) = \sum_{r=0}^{2n} b_r (x-3)^r$, we know that $b_n$ is the coefficient of $(x-3)^n$. According to Taylor's theorem, the coefficient $b_n$ is given by: $b_n = \frac{P^{(n)}(3)}{n!}$

Now, let's use the first representation of $P(x)$ to find its $n$-th derivative and evaluate it at $x=3$. $P(x) = \sum_{r=0}^{2n} a_r (x-2)^r$

We can split the sum based on the given condition $a_k=1$ for $k \ge n$: $P(x) = \sum_{r=0}^{n-1} a_r (x-2)^r + \sum_{r=n}^{2n} a_r (x-2)^r$ Since $a_r = 1$ for $r \ge n$: $P(x) = \sum_{r=0}^{n-1} a_r (x-2)^r + \sum_{r=n}^{2n} (x-2)^r$

Now, we need to find the $n$-th derivative of $P(x)$, $P^{(n)}(x)$. Consider the terms in the first sum: $\sum_{r=0}^{n-1} a_r (x-2)^r$. If $r < n$, then the $n$-th derivative of $(x-2)^r$ will be zero. For example, $\frac{d^n}{dx^n} (x-2)^0 = 0$, $\frac{d^n}{dx^n} (x-2)^1 = 0$, ..., $\frac{d^n}{dx^n} (x-2)^{n-1} = 0$. So, the first sum contributes zero to $P^{(n)}(x)$.

Thus, $P^{(n)}(x)$ is determined solely by the second sum: $P^{(n)}(x) = \frac{d^n}{dx^n} \left( \sum_{r=n}^{2n} (x-2)^r \right)$ $P^{(n)}(x) = \sum_{r=n}^{2n} \frac{d^n}{dx^n} (x-2)^r$

The $n$-th derivative of $(x-c)^r$ is given by $\frac{r!}{(r-n)!} (x-c)^{r-n}$. So, for $(x-2)^r$: $\frac{d^n}{dx^n} (x-2)^r = \frac{r!}{(r-n)!} (x-2)^{r-n}$

Substituting this back into $P^{(n)}(x)$: $P^{(n)}(x) = \sum_{r=n}^{2n} \frac{r!}{(r-n)!} (x-2)^{r-n}$

Now, evaluate $P^{(n)}(x)$ at $x=3$: $P^{(n)}(3) = \sum_{r=n}^{2n} \frac{r!}{(r-n)!} (3-2)^{r-n}$ $P^{(n)}(3) = \sum_{r=n}^{2n} \frac{r!}{(r-n)!} (1)^{r-n}$ $P^{(n)}(3) = \sum_{r=n}^{2n} \frac{r!}{(r-n)!}$

Finally, substitute this into the formula for $b_n$: $b_n = \frac{P^{(n)}(3)}{n!} = \frac{1}{n!} \sum_{r=n}^{2n} \frac{r!}{(r-n)!}$ $b_n = \sum_{r=n}^{2n} \frac{r!}{n!(r-n)!}$ Recognizing the definition of a binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$: $b_n = \sum_{r=n}^{2n} \binom{r}{n}$

This sum is a known binomial identity, often called the Hockey-stick identity or Christmas stocking identity: $\sum_{k=r}^{m} \binom{k}{r} = \binom{m+1}{r+1}$ In our case, the summation variable is $r$, the lower index of the binomial coefficient is $n$, and the upper limit of the summation is $2n$. Applying the identity: $b_n = \binom{2n+1}{n+1}$

Thus, we have shown that $b_n = ^{2n+1}C_{n+1}$.