1 câu trả lời
Giải thích các bước giải:
Ta có :
$a_k=C^k_{20}.3^k.(-5)^{20-k}$
$\to$Để $a_k$ lớn nhất
$\to a_k>0\to k=2q, 0\le q\le 10$
$\to a_{2q}=C^{2q}_{20}3^{2q}.5^{20-2q}$
$\to a_{2q}$ max
$\to a_{2q}\ge a_{2q-2}$ và $a_{2q}\ge a_{2q+2}$
+) $a_{2q}\ge a_{2q-2}$
$\to C^{2q}_{20}3^{2q}.5^{20-2q}\ge C^{2q-2}_{20}.3^{2q-2}.5^{20-2q+2}$
$\to\dfrac{20!}{(2q)!(20-2q)!}.9\ge \dfrac{20!}{(2q-2)!.(20-2q+2)!}.25$
$\to\dfrac{9}{(2q-2)(2q-1)}\ge \dfrac{25}{(20-2q+2)(20-2q+1)}$
$\to 1<q\le 4$
+) $a_{2q}\ge a_{2q+2}$
$\to C^{2q}_{20}3^{2q}.5^{20-2q}\ge C^{2q+2}_{20}.3^{2q+2}.5^{20-2q-2}$
$\to \dfrac{20!}{(2q)!.(20-2q)!}.25\ge \dfrac{20!}{(2q+2)!.(20-2q-2)!}.9$
$\to \dfrac{25}{(20-2q)(20-2q-1)}\ge \dfrac{9}{(2q+2)(2q+1)}$
$\to 4\le q\le 9$
$\to q=4\to k=8$
$\to a_k=C^8_{20}.3^{8}.(-5)^{12}$ là hệ số lớn nhất