Question

13．（3x+2）¹⁵展開式中最大系數(shù)是第7項．

Answer 1

分析 利用二項式（3x+2）¹⁵展開式的通項公式T_r+1，設(shè)第r+1項的系數(shù)最大，得出不等式組$\left\{\begin{array}{l}{{C}_{15}^{r}{•2}^{r}{•3}^{15-r}{≥C}_{15}^{r+1}{•2}^{r+1}{•3}^{14-r}}\\{{C}_{15}^{r}{•2}^{r}{•2}^{15-r}{≥C}_{15}^{r-1}{•2}^{r-1}{•3}^{16-r}}\end{array}\right.$，求出r的值即可．

解答 解：二項式（3x+2）¹⁵展開式的通項公式為
T_r+1=${C}_{15}^{r}$•2^r•3^15-r•x^15-r，
∴第r+1項的系數(shù)為${C}_{15}^{r}$•2^r•3^15-r；
設(shè)第r+1項的系數(shù)最大，則有
$\left\{\begin{array}{l}{{C}_{15}^{r}{•2}^{r}{•3}^{15-r}{≥C}_{15}^{r+1}{•2}^{r+1}{•3}^{14-r}}\\{{C}_{15}^{r}{•2}^{r}{•2}^{15-r}{≥C}_{15}^{r-1}{•2}^{r-1}{•3}^{16-r}}\end{array}\right.$，
即$\left\{\begin{array}{l}{\frac{15!•3}{r!•（15-r）!}≥\frac{15!•2}{（r+1）!•（14-r）!}}\\{\frac{15!•2}{r!•（15-r）!}≥\frac{15!•3}{（r-1）!•（16-r）!}}\end{array}\right.$，
化簡得$\left\{\begin{array}{l}{\frac{3}{15-r}≥\frac{2}{r+1}}\\{\frac{2}{r}≥\frac{3}{16-r}}\end{array}\right.$，
解得$\left\{\begin{array}{l}{r≥\frac{27}{5}}\\{r≤\frac{32}{5}}\end{array}\right.$，
即$\frac{27}{5}$≤r≤$\frac{32}{5}$；
又r∈N，∴得r=6，
∴（3x+2）¹⁵展開式中最大系數(shù)是第7項．
故答案為：第7項．

點(diǎn)評 本題考查了二項式定理與二項展開式的通項公式應(yīng)用問題，屬于基礎(chǔ)題．