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Step-by-Step Solution
Step 1: Express the Binomial Expansion
We start by writing the given expression in a form that is easier to analyze:
(3 + 6x)^n \;=\; 3^n (1 + 2x)^n.
This transformation allows us to manage the binomial coefficients more cleanly.
Step 2: Condition for the 9th Term to Be the Greatest (for x = 3/2)
We are told that the 9th term is numerically the greatest when x = \frac{3}{2} . In the sequence of binomial expansion terms, this means:
T_{8} \;\le\; T_{9} \;\ge\; T_{10},
where T_{r+1} denotes the (r+1)^\text{th} term.
Step 3: Express These Terms
The general term in the expansion of (3 + 6x)^n is:
T_{r+1} = \binom{n}{r} \, 3^{\,n-r} \,(6x)^r.
Hence, the 8th, 9th, and 10th terms are:
T_{8} = \binom{n}{7} \, 3^{n-7} \,(6x)^7,
T_{9} = \binom{n}{8} \, 3^{n-8} \,(6x)^8,
T_{10} = \binom{n}{9} \, 3^{n-9} \,(6x)^9.
Step 4: Apply Inequalities T_8 \le T_9 \ge T_{10}
We impose the conditions:
T_8 \;\le\; T_9 \quad\text{and}\quad T_9 \;\ge\; T_{10}.
Substitute x = \frac{3}{2} and simplify. This leads to constraints on n . Through simplification, one obtains the system:
27(n - 7) \;\ge\; 72 \quad\text{and}\quad 9(n - 8) \;\le\; 27,
which simplifies to:
n \;\ge\; \frac{29}{3} \quad\text{and}\quad n \;\le\; 11.
The smallest integer n satisfying these inequalities is n_0 = 10.
Step 5: Compute the Coefficients for x^6 and x^3 (When n=10 )
For n_0 = 10 , the general term again is:
T_{r+1} = \binom{10}{r} \, 3^{10-r} \,(6x)^r.
We find the terms corresponding to x^6 and x^3 .
Coefficient of x^6 : This occurs when r = 6.
\binom{10}{6} \, 3^{4} \,(6)^6.
Coefficient of x^3 : This occurs when r = 3.
\binom{10}{3} \, 3^{7} \,(6)^3.
Step 6: Form the Ratio k
The ratio k is defined as the coefficient of x^6 divided by the coefficient of x^3 :
k
=
\frac{\binom{10}{6} \, 3^{4} \,(6)^6}{\binom{10}{3} \, 3^{7} \,(6)^3}.
We simplify this expression carefully:
\binom{10}{6} = \frac{10!}{6!\,4!},
\quad
\binom{10}{3} = \frac{10!}{3!\,7!}.
Hence,
k
=
\frac{\frac{10!}{6!\,4!} \, 3^4 \, 6^6}{\frac{10!}{3!\,7!} \, 3^7 \, 6^3}
\;=\;
\frac{7!\,3!}{6!\,4!}
\,\cdot\,
\frac{6^6}{6^3}
\,\cdot\,
\frac{3^4}{3^7}.
Combine like terms and simplify step-by-step. One eventually gets:
k = 14.
Step 7: Compute k + n_0
We have k = 14 and n_0 = 10 , giving:
k + n_0 = 14 + 10 \;=\; 24.
Final Answer
24
