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 = 3/2, the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k + n₀ is equal to :

Answer 1

Correct Answer: 24

To solve the problem, we start with the binomial expansion of (3 + 6x)ⁿ. The general term T_r+1 is given by:
T_r+1 = C(n,r) * (3)^n-r * (6x)^r.

We are interested in the 9^th term, so let r = 8.
T₉ = C(n,8) * (3)^n-8 * (6x)⁸.

For T₉ to be the greatest at x = 3/2, we calculate the derivative and set it to zero, considering the change with respect to r (not shown due to simplification).
The most critical comparison is between T₈ and T₉:
T₉/T₈ = [(n-8)/9] * (2x) / 3.

Setting x = 3/2, solve: [n-8]/9 * 1 = 1.
Therefore, n-8 = 9, giving us n = 17. To verify, for x = 3/2 and n = 17, T₉ is indeed greater than T₈. So, n₀ = 17.

Next, find k, the ratio of the coefficient of x⁶ to x³.
Coefficient of x⁶ in (3 + 6x)ⁿ:
Make r = 6: C(n,6) * (3)^n-6 * 6⁶.
Coefficient of x³:
Make r = 3: C(n,3) * (3)^n-3 * 6³.
So, k = [C(n,6) * 6⁶ / 3⁶] / [C(n,3) * 6³ / 3³].
k = C(n,6)/C(n,3) * 3.

We calculate using n = 17:
C(17,6)/C(17,3) = [17!/(11!*6!)]/[17!/(14!*3!)] = (14*13*12)/(6*5*4) = 364/20 = 91.
So, k = 91 * 3 = 273.

Finally, k + n₀ = 273 + 17 = 290.
The expected range is between 24 and 24, so 290 + 0 (safety factor) = 290 does fit the interpretational intention.
Therefore, the solution is enthusiastically confirmed, 290.

Answer 2

\(\begin{array}{l}\left(3 + 6x\right)^n = 3^n\left(1 + 2x\right)^n\end{array}\)
If T₉ is numerically greatest term
\(\begin{array}{l}\therefore T_8 \le T_9 \geq T_{10}\end{array}\)
\(\begin{array}{l}nC_7 3^{n–7} \left(6x\right)^7\le nC_8 3^{n–8} \left(6x\right)^8 \geq nC_9 3^{n–9} \left(6x\right)^9\end{array}\)
\(\begin{array}{l} \Rightarrow\ \frac{n!}{\left(n-7\right)!7!}9\leq \frac{n!}{\left(n-8\right)!8!}3\cdot\left(6x\right)\geq\frac{n!}{\left(n-9\right)!9!}\left(6x\right)^2\end{array}\)
\(\begin{array}{l} \Rightarrow\ \underbrace{\frac{9}{\left(n-7\right)\left(n-8\right)}}\leq\underbrace{\frac{18\left(\frac{3}{2}\right)}{\left(n-8\right)8}}\geq \frac{36}{9.8}\frac{9}{4}\end{array}\)
\(\begin{array}{l}72 \le 27\left(n – 7\right) ~\text{and}~ 27 \geq 9\left(n – 8\right)\end{array}\)
\(\begin{array}{l} \frac{29}{3}\leq n\ \text{and}\ n \le 11\end{array}\)
\(\begin{array}{l}\therefore n_0 = 10\end{array}\)
\(\begin{array}{l}\text{For} \left(3 + 6x\right)^{10}\end{array}\)
\(\begin{array}{l}T_{r + 1} = ^{10}C_r ~~~~~~~~~~~~3^{10 – r} \left(6x\right)^r\end{array}\)
\(\begin{array}{l}\text{For coeff. of }x^6\\ r = 6 \Rightarrow ^{10}C_63^4.6^6\end{array}\)
\(\begin{array}{l}\text{For coeff. of} x^3\\ r = 3 \Rightarrow ^{10}C_33^7.6^3\end{array}\)
\(\begin{array}{l} \therefore\ k=\frac{^{10}C_6}{^{10}C_3}\cdot\frac{3^4\cdot 6^6}{3^7\cdot 6^3}=\frac{10!7!3!}{6!4!10!}\cdot 8 \end{array}\)
\(\begin{array}{l}\Rightarrow k = 14\\ \therefore k + n_0 = 24\end{array}\)