Question

If \[ \sum_{r=0}^{2n}a_{r}(x-2)^{r}=\sum_{r=0}^{2n}b_{r}(x-3)^{r} \] and \(a_{k}=1\ \forall k\ge1\), then the value of \[ \frac{b_{n}}{{}^{2n+1}C_{n+1}} \] is:

• A. $\frac{1}{2}$
• B. 2
• C. $\frac{1}{4}$
• D. 1

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!

Answer 1

The Correct Option is D

Concept: This shift relation is solved by changing variables to align the polynomial bases. By setting $y = x - 3$, the left-hand expression becomes a function of $(y+1)$, which can be expanded using the Binomial Theorem to compare coefficients against the right-hand polynomial. Step 1: Apply variable shift to align the bases.
Let us substitute $y = x - 3$ into the identity equation. This implies that $x - 2 = y + 1$. Substituting these back into the given expression yields: \[ \sum_{r=0}^{2n} a_r (y + 1)^r = \sum_{r=0}^{2n} b_r y^r \quad \cdots (1) \] We are given that $a_k = 1$ for all $k \ge 1$. Let us split the left-hand summation to separate the $a_0$ term: \[ a_0 + \sum_{r=1}^{2n} 1 \cdot (y + 1)^r = \sum_{r=0}^{2n} b_r y^r \quad \cdots (2) \]

Step 2: Isolate the coefficient of $y^n$ using binomial expansion.
We need to find the value of $b_n$, which represents the coefficient of $y^n$ on the right side of equation (2). Let us extract the coefficient of $y^n$ from the left side:
• The constant term $a_0$ does not contain any $y$ powers, so its contribution to $y^n$ is 0.
• For the remaining summation $\sum_{r=1}^{2n} (y + 1)^r$, each term $(y + 1)^r$ can be expanded using the general binomial formula: $(y+1)^r = \sum_{k=0}^{r} {}^{r}C_k y^k$.
• The term $y^n$ will appear in every index block where $r \ge n$. Its coefficient in any specific block $r$ is exactly given by the combination ${}^{r}C_n$. Summing these contributions from $r = n$ to $2n$ gives the total coefficient: \[ b_n = \sum_{r=n}^{2n} {}^{r}C_n = {}^{n}C_n + {}^{n+1}C_n + {}^{n+2}C_n + \dots + {}^{2n}C_n \quad \cdots (3) \]

Step 3: Simplify the combination sum using the Hockey-Stick Identity.
Recall the Hockey-Stick Identity for binomial coefficients: \[ \sum_{r=n}^{m} {}^{r}C_n = {}^{m+1}C_{n+1} \] Applying this identity directly to equation (3) by setting the upper limit parameter to $m = 2n$: \[ b_n = {}^{2n+1}C_{n+1} \]

Step 4: Evaluate the final target ratio fraction.
Substitute this simplified expression for $b_n$ back into our target parameter equation: \[ \text{Target Value} = \frac{b_n}{{}^{2n+1}C_{n+1}} = \frac{{}^{2n+1}C_{n+1}}{{}^{2n+1}C_{n+1}} = 1 \] This yields exactly 1, matching option (D) perfectly.