$C^{2}_{n}.C^{n-2}_{n} + 2C^{n-2}_{n}.C^{n-1}_{n} + C^{1}_{n}.C^{n-1}_{n} = 11025$ ĐK: $n\in N^*,n\geq 2$<=> $\frac{n!.n!}{2!.(n-2)!.(n-2)!.2!} + \frac{2n!.n!}{(n-2)!.(n-1)} + \frac{n!.n!}{(n-1)!.(n-1)!} = 11025$
<=> $n^2.(n+1)^2=44100$
<=> $n^2 + n - 210 = 0$
<=>$\left[ {} \right.\begin{matrix} n=14\\n=-15(loại)\end{matrix}$
Ta có $(\frac{1}{2}+\frac{x}{3})^{14}$ = $\sum_{k=0}^{14}C^{k}_{14}(\frac{1}{2})^{14-k}.(\frac{1}{3})^k.x^k$
Suy ra $a_{k}$= $C^{k}_{14}.(\frac{1}{2})^{14-k}.(\frac{1}{3})^k$
$a_{k}\geq a_{k+1}$ <=> $C^{k}_{14}.(\frac{1}{2})^{14-k}.(\frac{1}{3})^k\geq C^{k+1}_{14}.(\frac{1}{2})^{13-k}.(\frac{1}{3})^{k+1}$
<=> $\frac{14!}{k!.(14-k)!}.(\frac{1}{2})^{14-k}.(\frac{1}{3})^k\geq \frac{14!}{(k+1)!.(13-k)!}.(\frac{1}{2})^{13-k}.(\frac{1}{3})^{k+1}$
<=> $3k+3\geq 28-2k$ <=> $k\geq 5$
Vậy ta có:$\begin{cases}a_{6}\geq a_{7}\geq ...\geq a_{14}\\ a_{1}\leq a_{2}\leq ...\leq a_{5} \end{cases}$
Mà ta lại có $a_{5}=a_{6}$
Vậy hệ số lớn nhất là $a_{5} và a_{6}$