Let the coefficients of three consecutive terms T_r, T_{r+1} and T_{r+2} in the binomial expansion o

Q: Wait, why did we get $13=0$ in the first part? Did I make a mistake?

You did not make a mistake! This is a classic JEE trap. The algebra leads to a contradiction, which simply means that there is no integer $r$ that satisfies the condition. It is a powerful way of proving that such terms do not exist, so $p=0$ is the correct conclusion.

Q: Why is $k$ a multiple of 12 in the second part?

Great question! For the term to be rational, the exponents of the radicals must be integers. We need $\frac{12-k}{4} \in \mathbb{Z}$ and $\frac{k}{3} \in \mathbb{Z}$. This implies $k$ must be a multiple of 4 and $k$ must be a multiple of 3. Therefore, $k$ must be a multiple of their Least Common Multiple, which is $\text{LCM}(4, 3) = 12$.

Q: Are there any other rational terms besides $k=0$ and $k=12$?

We checked all possible values of $k$ from 0 to 12. Since $k$ must be a multiple of 12, only $k=0$ and $k=12$ are valid within the range $[0, 12]$. There are no other possibilities.

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 $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $p+q$ is equal to:

Options:

(A) 283

(B) 287

(C) 295

(D) 299

Sigma Percentile · Accepted Answer

### Analyzing the G.P. Coefficients

We begin with the expansion of $(a+b)^{12}$. We seek three consecutive terms $T_r, T_{r+1}, T_{r+2}$ whose coefficients form a Geometric Progression.

Let the coefficients be $C_1 = ^{12}C_{r-1}$, $C_2 = ^{12}C_r$, and $C_3 = ^{12}C_{r+1}$. The condition for these to be in G.P. is $C_2^2 = C_1 \cdot C_3$, which implies:

$$(^{12}C_r)^2 = ^{12}C_{r-1} \cdot ^{12}C_{r+1}$$

Using the definition $^{n}C_r = \frac{n!}{r!(n-r)!}$, we substitute into the equation:

$$\left( \frac{12!}{r!(12-r)!} ight)^2 = \frac{12!}{(r-1)!(13-r)!} \cdot \frac{12!}{(r+1)!(11-r)!}$$

### The Master Equation

The $(12!)^2$ terms on both sides cancel out completely. By rearranging the remaining factorials, we obtain the following ratio:

$$\frac{(r+1)!(11-r)!}{r!(12-r)!} = \frac{(12-r)!(r-1)!}{(13-r)!(r+1)!} 	ext{ (simplified to) } \frac{r+1}{12-r} = \frac{13-r}{r}$$

Cross-multiplying gives us $(r+1)r = (13-r)(12-r)$. Expanding both sides, we get:

$$r^2 + r = 156 - 25r + r^2$$

The $r^2$ terms vanish, leaving $26r = 156$, which yields $r = 6$. Since a valid integer $r$ exists, the coefficients of the 6th, 7th, and 8th terms form a G.P. Thus, $p=1$ (representing the existence of such terms).

### The Rationality Quest

We now find the sum of rational terms in the expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$, which we rewrite as $(3^{\frac{1}{4}} + 4^{\frac{1}{3}})^{12}$. The general term is:

$$T_{k+1} = ^{12}C_k (3^{\frac{1}{4}})^{12-k} (4^{\frac{1}{3}})^k = ^{12}C_k \cdot 3^{\frac{12-k}{4}} \cdot 4^{\frac{k}{3}}$$

For the term to be rational, the exponents must be integers. This requires $\frac{12-k}{4} \in \mathbb{Z}$ and $\frac{k}{3} \in \mathbb{Z}$.

This implies $k$ must be a multiple of 3, and $(12-k)$ must be a multiple of 4. In the range $0 \le k \le 12$, the possible values for $k$ are $0, 3, 6, 9, 12$. Checking the conditions:

1. For $k=0$: $T_1 = ^{12}C_0 \cdot 3^3 \cdot 4^0 = 27$ (Rational)
2. For $k=12$: $T_{13} = ^{12}C_{12} \cdot 3^0 \cdot 4^4 = 256$ (Rational)

### Final Calculation

The sum of the rational terms is $q = 27 + 256 = 283$.

Given our previous finding that such terms exist ($p=1$), the final result is:

**$p + q = 1 + 283 = 284$**

Select Answer:

Visualized Solution

Problem Overview

Identifying Coefficients

Applying G.P. Condition

Expanding the G.P. Equation

Simplifying Factorials

Solving for $r$

Conclusion for $p$

Second Expansion Analysis

General Term $T_{k + 1}$

Rationality Constraint

Finding Values of $k$

Calculating Term for $k = 0$

Calculating Term for $k = 12$

Final Result $p + q$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the G.P

The Master Equation

The Rationality Quest

Final Calculation

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Select Answer:

Visualized Solution

Problem Overview

Identifying Coefficients

Applying G.P. Condition

Expanding the G.P. Equation

Simplifying Factorials

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Second Expansion Analysis

General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk+1​

Rationality Constraint

Finding Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Calculating Term for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=0

Calculating Term for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=12

Final Result .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p+q