If
\[
\sum_{r=0}^{2n}a_{r}(x-2)^{r}=\sum_{r=0}^{2n}b_{r}(x-3)^{r}
\]
and \(a_{k}=1\ \forall k\ge1\), then the value of
\[
\frac{b_{n}}{{}^{2n+1}C_{n+1}}
\]
is:
Show Hint
To solve this quickly, try testing the minimal case $n=1$. The equation expands to $a_0 + a_1(x-2) + a_2(x-2)^2 = b_0 + b_1(x-3) + b_2(x-3)^2$. Setting $a_1=a_2=1$ and expanding the quadratics gives a coefficient of $b_1 = 3$. The denominator becomes ${}^3C_2 = 3$. The ratio is $3/3 = 1$, confirming choice (D) instantly!
Step 1: Understanding the Concept:
This problem involves shifting the base of a polynomial power series. We are given the coefficients for an expansion centered at \( x=2 \) and asked for a specific coefficient centered at \( x=3 \). Step 2: Key Formula or Approach:
1. Let \( y = x - 3 \). Then \( x - 2 = y + 1 \).
2. Rewrite the summation: \( \sum a_r(y + 1)^r = \sum b_r y^r \).
3. Use the binomial theorem to expand \( (y + 1)^r \). Step 3: Detailed Explanation:
Given \( a_k = 1 \) for \( k \ge 1 \), and assuming \( a_0 = 1 \) for a consistent GP (though \( a_0 \) doesn't affect \( b_n \) for \( n \ge 1 \)).
LHS: \( \sum_{r=0}^{2n} (y + 1)^r = \frac{(y + 1)^{2n+1} - 1}{(y + 1) - 1} = \frac{(y + 1)^{2n+1} - 1}{y} \).
We need to find \( b_n \), which is the coefficient of \( y^n \) in the expansion of this expression.
The numerator is \( (y + 1)^{2n+1} - 1 = \left[ \binom{2n+1}{0}y^0 + \binom{2n+1}{1}y^1 + \dots + \binom{2n+1}{n+1}y^{n+1} + \dots \right] - 1 \).
The first term \( \binom{2n+1}{0} = 1 \) cancels with the \( -1 \).
Numerator \( = \binom{2n+1}{1}y^1 + \binom{2n+1}{2}y^2 + \dots + \binom{2n+1}{n+1}y^{n+1} + \dots \).
Dividing by \( y \):
LHS \( = \binom{2n+1}{1}y^0 + \binom{2n+1}{2}y^1 + \dots + \binom{2n+1}{n+1}y^n + \dots \).
The coefficient of \( y^n \) is \( b_n = \binom{2n+1}{n+1} \).
The question asks for the ratio:
\[ \frac{b_n}{^{2n+1}C_{n+1}} = \frac{^{2n+1}C_{n+1}}{^{2n+1}C_{n+1}} = 1 \]. Step 4: Final Answer:
The ratio is 1, corresponding to option (D).