Algebra – Binomial Theorem – JEE Main 28 Jan 2025 Shift 2

Solution:

Part 1: Find $p$.
The coefficients are $^{12}C_{r-1}, ^{12}C_r, ^{12}C_{r+1}$ (for terms $T_r, T_{r+1}, T_{r+2}$).
Since they are in G.P.:
$$(^{12}C_r)^2 = ^{12}C_{r-1} \times ^{12}C_{r+1}$$

$$\frac{^{12}C_r}{^{12}C_{r-1}} = \frac{^{12}C_{r+1}}{^{12}C_r}$$

Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:
$$\frac{12 – r + 1}{r} = \frac{12 – (r+1) + 1}{r+1}$$
$$\frac{13-r}{r} = \frac{12-r}{r+1}$$
$$(13-r)(r+1) = r(12-r)$$
$$13r + 13 – r^2 – r = 12r – r^2$$
$$12r + 13 – r^2 = 12r – r^2$$
$$13 = 0$$
This is impossible. Thus, no such $r$ exists.
So, $p = 0$.

Part 2: Find $q$.
Expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12} = (3^{1/4} + 4^{1/3})^{12}$.
General term $T_{k+1} = ^{12}C_k (3^{1/4})^{12-k} (4^{1/3})^k$.
For terms to be rational, the powers of 3 and 4 must be integers.
Power of 3: $\frac{12-k}{4}$ must be integer $\Rightarrow 12-k \in \{0, 4, 8, 12\} \Rightarrow k \in \{0, 4, 8, 12\}$.
Power of 4: $\frac{k}{3}$ must be integer $\Rightarrow k \in \{0, 3, 6, 9, 12\}$.

Common values of $k$ are $0$ and $12$.

For $k=0$:
Term $= ^{12}C_0 (3^{12/4}) (4^0) = 1 \times 3^3 \times 1 = 27$.

For $k=12$:
Term $= ^{12}C_{12} (3^0) (4^{12/3}) = 1 \times 1 \times 4^4 = 256$.

Sum of rational terms $q = 27 + 256 = 283$.

Final Answer:
$p + q = 0 + 283 = 283$.

Ans. (1)