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

Question

Let for the 9th term in the binomial expansion of $(3+6x)^n$, in the increasing powers of $6x$, to be the greatest for $x=\dfrac32$, the least value of $n$ be $n_0$. If $k$ is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$, then $k+n_0$ is equal to:

Answer: 24

Accepted Answer

$(3+6x)^n=3^n(1+2x)^n$. For $T_9$ to be numerically greatest at $x=3/2$ (so $2x=3$): $T_8\le T_9\ge T_{10}$, giving $\dfrac{29}{3}\le n\le11$, so the least integer is $n_0=10$.

For $(3+6x)^{10}$: coefficient of $x^6$ ($r=6$): $\binom{10}{6}3^4\,6^6$; coefficient of $x^3$ ($r=3$): $\binom{10}{3}3^7\,6^3$.

$$k=\frac{\binom{10}{6}3^4\,6^6}{\binom{10}{3}3^7\,6^3}=\frac{7!3!}{6!4!}\times8=14$$

$k+n_0=14+10=24$

Solutions · 1