浙江 2020 · 数学 q12

题面

12．设$( 1 + 2 x ) \^{5} = a_{1} + a_{2} x + a_{3} x^{2} + a_{4} x^{3} + a_{5} x^{4} + a_{6} x^{5}$，则 a₅＝$a_{1} + a_{2} + a_{3} =$

解析

${\left( {1 + 2 x} \right)}^{5}$的通项为$T_{r + 1} = C_{5}^{r} ( 2 x )^{r} = 2^{r} C_{5}^{r} x^{r}$令$r = 4$，则$T_{5} = 2^{4} C_{5}^{4} x^{4} = 8 0 x^{4}$，故a_{5} = 8 0$$a_{1} + a_{3} + a_{5} = 2^{1} C_{5}^{1} + 2^{3} C_{5}^{3} + 2^{5} C_{5}^{5} = 1 2 2