Let the 9^{text{th}} term in the binomial exp | vedclass

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(A) The general term is $T_{r+1} = {}^{n}C_{r} 3^{n-r} (6x)^{r} = {}^{n}C_{r} 3^{n-r} 6^{r} x^{r}$.For $x = \frac{3}{2}$,$T_{r+1} = {}^{n}C_{r} 3^{n-r

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) The general term is $T_{r+1} = {}^{n}C_{r} 3^{n-r} (6x)^{r} = {}^{n}C_{r} 3^{n-r} 6^{r} x^{r}$.For $x = \frac{3}{2}$,$T_{r+1} = {}^{n}C_{r} 3^{n-r} 6^{r} (\frac{3}{2})^{r} = {}^{n}C_{r} 3^{n-r} 3^{r} 2^{r} \frac{3^{r}}{2^{r}} = {}^{n}C_{r} 3^{n+r}$.Since $T_{9}$ is the greatest term,$T_{9} \ge T_{10}$ and $T_{9} \ge T_{8}$.From $\frac{T_{9}}{T_{10}} \ge 1$,we have $\frac{{}^{n}C_{8} 3^{n+8}}{{}^{n}C_{9} 3^{n+9}} \ge 1 \implies \frac{9}{n-8} \cdot \frac{1}{3} \ge 1 \implies 3 \ge n-8 \implies n \le 11$.From $\frac{T_{9}}{T_{8}} \ge 1$,we have $\frac{{}^{n}C_{8} 3^{n+8}}{{}^{n}C_{7} 3^{n+7}} \ge 1 \implies \frac{n-7}{8} \cdot 3 \ge 1 \implies 3n-21 \ge 8 \implies 3n \ge 29 \implies n \ge 9.66$.Thus,the least integer $n_{0} = 10$.For $n=10$,the expansion is $(3+6x)^{10}$.The coefficient of $x^{6}$ is ${}^{10}C_{6} 3^{4} 6^{6} = 210 \cdot 3^{4} \cdot 2^{6} \cdot 3^{6} = 210 \cdot 3^{10} \cdot 2^{6}$.The coefficient of $x^{3}$ is ${}^{10}C_{3} 3^{7} 6^{3} = 120 \cdot 3^{7} \cdot 2^{3} \cdot 3^{3} = 120 \cdot 3^{10} \cdot 2^{3}$.The ratio $k = \frac{210 \cdot 3^{10} \cdot 2^{6}}{120 \cdot 3^{10} \cdot 2^{3}} = \frac{210}{120} \cdot 2^{3} = \frac{7}{4} \cdot 8 = 14$.Therefore,$k + n_{0} = 14 + 10 = 24$.

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