Let for the 9ᵗʰ term in the binomial expansion of (3+6x)ⁿ, in the increasing powers of 6x, to be the

Q: Why do we need two inequalities? Can't we just check one?

Think of a mountain peak. If you only check the left side, you know you are higher than the previous step, but you might still be climbing! To be the absolute peak, you must be higher than the previous step AND higher than the next step. Mathematically, $T_9 \ge T_8$ ensures we aren't still increasing, and $T_9 \ge T_{10}$ ensures we aren't already decreasing.

Q: Why is $n$ restricted to integers?

In the binomial expansion $(a+b)^n$, $n$ represents the power of the expression. By definition, the power in a standard binomial expansion is a non-negative integer. When we solve the inequality $9.66 \le n \le 11$, we are looking for the valid integer values that satisfy the condition. Since $n$ must be an integer, we select from the set $\{10, 11\}$.

Q: Is there a faster way to calculate $k$ without expanding the combinations?

Absolutely! Never calculate the full value of $\binom{10}{6}$ or $\binom{10}{3}$ if you can avoid it. Use the property $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and simplify the ratio algebraically. You will find that most terms cancel out, leaving you with a much simpler expression. This is the 'JEE way'—work smarter, not harder.

Question

Let for the $9^{th}$ term in the binomial expansion of $(3+6x)^n$, in the increasing powers of $6x$, to be the greatest for $x = \frac{3}{2}$, the least value of $n$ is $n_0$. If $k$ is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$, then $k + n_0$ is equal to:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the summit of the binomial expansion $(3+6x)^n$, we treat the terms as peaks in a range. We are given that at $x = \frac{3}{2}$, the $9^{th}$ term, $T_9$, is the greatest term.

The condition for $T_9$ to be the greatest term is defined by the sandwich inequality:
$$T_9 \ge T_8 \quad 	ext{and} \quad T_9 \ge T_{10}$$

### The Sandwich Inequality

We utilize the ratio formula for consecutive terms in a binomial expansion $(a+b)^n$:
$$\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot \frac{b}{a}$$

Given $a=3$ and $b=6x$, at $x = \frac{3}{2}$, the ratio $\frac{b}{a}$ becomes:
$$\frac{b}{a} = \frac{6(\frac{3}{2})}{3} = 3$$

For the first condition, $T_9 \ge T_8$, we set $r=8$:
$$\frac{n-8+1}{8} \cdot 3 \ge 1$$
$$3(n-7) \ge 8 \implies 3n - 21 \ge 8 \implies 3n \ge 29 \implies n \ge 9.66$$

For the second condition, $T_9 \ge T_{10}$, we use the ratio $\frac{T_{10}}{T_9} \le 1$ by setting $r=9$:
$$\frac{n-9+1}{9} \cdot 3 \le 1$$
$$\frac{n-8}{3} \le 1 \implies n - 8 \le 3 \implies n \le 11$$

Combining these results, we find $9.66 \le n \le 11$. Since $n$ must be an integer, $n$ can be $10$ or $11$. The least value is **$n_0 = 10$**.

### The Elegance of Ratios

With $n=10$, we calculate $k$, the ratio of the coefficient of $x^6$ to the coefficient of $x^3$. The general term is $T_{r+1} = \binom{10}{r} 3^{10-r} (6x)^r$.

The coefficient of $x^6$ is $\binom{10}{6} 3^4 6^6$, and the coefficient of $x^3$ is $\binom{10}{3} 3^7 6^3$. The ratio $k$ is:
$$k = \frac{\binom{10}{6} 3^4 6^6}{\binom{10}{3} 3^7 6^3}$$

Simplifying the binomial coefficients and powers:
$$\frac{\binom{10}{6}}{\binom{10}{3}} = \frac{210}{120} = \frac{7}{4}$$
$$\frac{3^4}{3^7} = \frac{1}{27}, \quad \frac{6^6}{6^3} = 6^3 = 216$$

Combining these values:
$$k = \frac{7}{4} \cdot \frac{1}{27} \cdot 216 = \frac{7}{4} \cdot 8 = 14$$

### Final Calculation

The final result is the sum of the least value $n_0$ and the ratio $k$:
$$k + n_0 = 14 + 10 = \mathbf{24}$$

Enter Numerical Value:

Visualized Solution

General Term of the Expansion

Condition for Greatest Term: $T_{9} \geq T_{8}$

Setting up the First Inequality

Solving the First Inequality

Condition for Greatest Term: $T_{9} \geq T_{10}$

Solving the Second Inequality

Finding the Least Value $n_{0}$

Setting up the Ratio $k$

Extracting the Coefficients

Simplifying the Ratio $k$

Final Calculation: $k + n_{0}$

The Sigma Insight: Greatest Term in Expansion

Analyzing the Setup

The Sandwich Inequality

The Elegance of Ratios

Final Calculation