Question

If \( 26 \left( \dfrac{2^3 \cdot \binom{12}{2}}{3} + \dfrac{2^5 \cdot \binom{12}{4}}{5} + \dots + \dfrac{2^{13} \cdot \binom{12}{12}}{13} \right) = 3^{13 - \alpha} \), then find the value of \( \alpha \).

Answer 1

Correct Answer: 25

Step 1: Starting from the Binomial Theorem:
We begin with the binomial expansion: \[ (1+x)^{12} = \sum_{r=0}^{12}\binom{12}{r}x^r = \binom{12}{0} + \binom{12}{1}x + \binom{12}{2}x^2 + \cdots + \binom{12}{12}x^{12} \] Step 2: Integrating Both Sides from $-2$ to $2$:
Integrating term by term: \[ \left[\frac{(1+x)^{13}}{13}\right]_{-2}^{2} = \left[\binom{12}{0}x + \binom{12}{1}\frac{x^2}{2} + \binom{12}{2}\frac{x^3}{3} + \cdots + \binom{12}{12}\frac{x^{13}}{13}\right]_{-2}^{2} \] Step 3: Evaluating the Left Side:
\[ \frac{(1+2)^{13}}{13} - \frac{(1-2)^{13}}{13} = \frac{3^{13}}{13} - \frac{(-1)^{13}}{13} = \frac{3^{13}}{13} + \frac{1}{13} = \frac{3^{13}+1}{13} \] Step 4: Evaluating the Right Side:
The right side at $x=2$ minus at $x=-2$. Because the polynomial has both even and odd powers, subtracting $x=-2$ from $x=2$ eliminates even powers and doubles odd powers: \[ \text{RHS} = 2\left[\binom{12}{0}\cdot 2 + \binom{12}{2}\cdot\frac{2^3}{3} + \binom{12}{4}\cdot\frac{2^5}{5} + \cdots + \binom{12}{12}\cdot\frac{2^{13}}{13}\right] \] Step 5: Separating the $\binom{12{0}$ Term:}
\[ \frac{3^{13}+1}{13} = 2\left[2\cdot\binom{12}{0} + \binom{12}{2}\cdot\frac{2^3}{3} + \cdots + \binom{12}{12}\cdot\frac{2^{13}}{13}\right] \] \[ 3^{13}+1 = 26\left[2 + \binom{12}{2}\cdot\frac{2^3}{3} + \binom{12}{4}\cdot\frac{2^5}{5} + \cdots + \binom{12}{12}\cdot\frac{2^{13}}{13}\right] \] \[ 3^{13}+1 = 26\cdot 2 + 26\left[\binom{12}{2}\cdot\frac{2^3}{3} + \binom{12}{4}\cdot\frac{2^5}{5} + \cdots + \binom{12}{12}\cdot\frac{2^{13}}{13}\right] \] \[ 3^{13}+1 - 52 = 26\left[\frac{2^3\binom{12}{2}}{3} + \frac{2^5\binom{12}{4}}{5} + \cdots + \frac{2^{13}\binom{12}{12}}{13}\right] \] \[ 3^{13} - 51 = 26\left[\frac{2^3\cdot{}^{12}C_2}{3} + \frac{2^5\cdot{}^{12}C_4}{5} + \cdots + \frac{2^{13}\cdot{}^{12}C_{12}}{13}\right] \] Step 6: Identifying $\alpha$:
Comparing with $3^{13} - \alpha$: \[ \alpha = 51 \] Step 7: Final Answer:
\[ \alpha = 51 \]