Question

Let the coefficients of three consecutive terms \( T_r \), \( T_{r+1} \), and \( T_{r+2} \) in the binomial expansion of \( (a + b)^{12}\) be in a G.P. and let \( p \) be the number of all possible values of \( r \). Let \( q \) be the sum of all rational terms in the binomial expansion of \( \left( 4\sqrt{3} + 3\sqrt{4} \right)^{12} \). Then \( p + q \) is equal to:

• A. 283
• B. 295
• C. 287
• D. 299

Hint:

When dealing with G.P. relations in binomial expansions, equate the ratio of the coefficients of consecutive terms to derive relationships between the terms.

Answer 1

The Correct Option is A

Step 1: Identify Binomial Expansion Terms

The binomial expansion of \( (a + b)^{12} \) yields terms in the format: \[ T_r = \binom{12}{r} a^{12-r} b^r \] The coefficients of three consecutive terms, \( T_r \), \( T_{r+1} \), and \( T_{r+2} \), are stated to form a geometric progression (G.P.).

Step 2: Form the Ratio Equation

The condition for a G.P. is expressed as: \[ \frac{T_{r+1}}{T_r} = \frac{T_{r+2}}{T_{r+1}} \] Substituting the binomial coefficients gives: \[ \frac{\binom{12}{r+1}}{\binom{12}{r}} = \frac{\binom{12}{r+2}}{\binom{12}{r+1}} \] This simplifies to: \[ \frac{12-r}{r+1} = \frac{12-r-1}{r+2} \]

Step 3: Solve the Quadratic Equation

Expanding and simplifying the equation: \[ 13 - r = 12r - r^2 \] Rearranging the terms: \[ 13 = r(12 - r) \] This further simplifies to: \[ 13 = 12r - r^2 \] Solving this quadratic equation yields no valid values for \( r \), thus setting \( p = 0 \).

Step 4: Calculate the Sum of Rational Terms

For the expansion of \( \left( 4\sqrt{3} + 3\sqrt{4} \right)^{12} \), the general term is: \[ T_r = \binom{12}{r} (4\sqrt{3})^{12-r} (3\sqrt{4})^r \] Rational terms occur when the exponents of the square roots are even. The sum of these rational terms is calculated as: \[ q = 27 + 256 = 283 \] Therefore, the final sum is: \[ p + q = 0 + 283 = 283 \]