Question

If \(\alpha_n\) is the coefficient of \(x^n\) in the expansion of \((1-x)^{-5}\) and \(\beta_n\) is the coefficient of \(x^n\) in the expansion of \((1-x)^{-4}\), then \(\alpha_{12} + \beta_{13}\) is equal to: 

• A. \(\alpha_{13}\)
• B. \(\beta_{13}\)
• C. \(25\)
• D. \(\beta_{25}\)

Hint:

The coefficient of \(x^n\) in \((1-x)^{-r}\) is \[ {}^{n+r-1}C_{r-1}. \] After obtaining coefficients, look for Pascal's identity to simplify expressions involving sums of combinations.

Answer 1

The Correct Option is A

Concept: For the expansion \[ (1-x)^{-r}, \] the coefficient of \(x^n\) is \[ {}^{n+r-1}C_{r-1}. \]

Step 1: Find \(\alpha_{12}\).
Since \[ (1-x)^{-5}, \] we have \[ \alpha_n = {}^{n+4}C_4. \] Thus, \[ \alpha_{12} = {}^{16}C_4. \]

Step 2: Find \(\beta_{13}\).
Since \[ (1-x)^{-4}, \] we obtain \[ \beta_n = {}^{n+3}C_3. \] Therefore, \[ \beta_{13} = {}^{16}C_3. \]

Step 3: Use Pascal's identity.
\[ \alpha_{12}+\beta_{13} = {}^{16}C_4+{}^{16}C_3. \] Using \[ {}^nC_r+{}^nC_{r-1} = {}^{n+1}C_r, \] we get \[ {}^{16}C_4+{}^{16}C_3 = {}^{17}C_4. \] But \[ \alpha_{13} = {}^{17}C_4. \] Hence, \[ \alpha_{12}+\beta_{13} = \alpha_{13}. \] \[ \boxed{\alpha_{13}} \]