Question

2．二項(xiàng)式（2x+3）¹²的展開式中系數(shù)最大的項(xiàng)的項(xiàng)數(shù)是：8．

Answer 1

分析 先求得二項(xiàng)式（2x+3）¹²的展開式的通項(xiàng)公式，可得第r+1項(xiàng)的系數(shù)為${C}_{12}^{r}$•3^r•2^12-r．設(shè)第r+1項(xiàng)的系數(shù)最大，則由 $\left\{\begin{array}{l}{{C}_{12}^{r}{•3}^{r}{•2}^{12-r}{≥C}_{12}^{r+1}{•3}^{r+1}{•2}^{11-r}}\\{{C}_{12}^{r}{•3}^{r}{{•2}^{12-r}≥C}_{12}^{r-1}{•3}^{r-1}{•2}^{13-r}}\end{array}\right.$求得r的范圍，從而得出結(jié)論．

解答 解：二項(xiàng)式（2x+3）¹²的展開式的通項(xiàng)公式為T_r+1=${C}_{12}^{r}$•3^r•2^12-r•x^12-r，
可得第r+1項(xiàng)的系數(shù)為${C}_{12}^{r}$•3^r•2^12-r．
設(shè)第r+1項(xiàng)的系數(shù)最大，則有 $\left\{\begin{array}{l}{{C}_{12}^{r}{•3}^{r}{•2}^{12-r}{≥C}_{12}^{r+1}{•3}^{r+1}{•2}^{11-r}}\\{{C}_{12}^{r}{•3}^{r}{{•2}^{12-r}≥C}_{12}^{r-1}{•3}^{r-1}{•2}^{13-r}}\end{array}\right.$，
即 $\left\{\begin{array}{l}{\frac{12！{•3}^{r}{•2}^{12-r}}{r!•（12-r）!}≥\frac{12!{•3}^{r+1}{•2}^{11-r}}{（r+1）!•（11-r）!}}\\{\frac{12!{•3}^{r}{•2}^{12-r}}{r!•（12-r）!}≥\frac{12!{•3}^{r-1}{•2}^{13-r}}{（r-1）!•（13-r）!}}\end{array}\right.$，即$\left\{\begin{array}{l}{r≥\frac{34}{5}}\\{r≤\frac{39}{5}}\end{array}\right.$，
求得$\frac{34}{5}$≤r≤$\frac{39}{5}$．
再結(jié)合r∈N，可得r=7，
故答案為：8．

點(diǎn)評(píng) 本題主要考查二項(xiàng)式定理的應(yīng)用，二項(xiàng)式展開式的通項(xiàng)公式的應(yīng)用，組合數(shù)的計(jì)算公式，屬于中檔題．