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

The binomial expansion of \( (a + b)^{12} \) gives terms of the form: \[ T_r = \binom{12}{r} a^{12-r} b^r \] We are given that the coefficients of three consecutive terms \( T_r \), \( T_{r+1} \), and \( T_{r+2} \) form a geometric progression (G.P.). So, the ratio of the coefficients of the terms should be equal: \[ \frac{T_{r+1}}{T_r} = \frac{T_{r+2}}{T_{r+1}} \] Using the binomial coefficients, this simplifies to: \[ \frac{\binom{12}{r+1}}{\binom{12}{r}} = \frac{\binom{12}{r+2}}{\binom{12}{r+1}} \] This results in the following equation: \[ \frac{12-r}{r+1} = \frac{12-r-1}{r+2} \] By simplifying, we obtain: \[ 13 - r = 12r - r^2 \quad \Rightarrow \quad 13 = r(12 - r) \] This simplifies to: \[ 13 = 12r - r^2 \] Solving this quadratic equation gives no valid values for \( r \), so \( p = 0 \). Next, for the sum of rational terms in the binomial expansion of \( \left( 4\sqrt{3} + 3\sqrt{4} \right)^{12} \), the general term is: \[ T_r = \binom{12}{r} (4\sqrt{3})^{12-r} (3\sqrt{4})^r \] The rational terms occur when the exponents of the square roots are even, so we consider the terms with even powers of \( 3 \) and \( 4 \). We calculate the sum of these rational terms: \[ q = 27 + 256 = 283 \] Thus, \( p + q = 0 + 283 = 283 \).