Cho \(P(x)={{\left( 1+3x+{{x}^{2}} \right)}^{20}}\...

D

- Hướng dẫn giải

Phương pháp giải:

Công thức nhị thức Newton: \({{\left( x+y \right)}^{n}}=C_{n}^{0}{{x}^{n}}+C_{n}^{1}{{x}^{n-1}}y+...+C_{n}^{n}{{y}^{n}}=\sum\limits_{i=0}^{n}{C_{n}^{i}{{x}^{n-i}}{{y}^{i}}}\)

Giải chi tiết:

\(\begin{align}  & P(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{40}}{{x}^{40}} \\ & \Rightarrow P'(x)={{a}_{1}}+2{{a}_{2}}x+...+40{{a}_{40}}{{x}^{39}} \\\end{align}\)

Ta có: \(P(x)={{\left( 1+3x+{{x}^{2}} \right)}^{20}}\Rightarrow P'(x)=20{{(1+3x+{{x}^{2}})}^{19}}(3+2x)\)

\(\Rightarrow 20{{(1+3x+{{x}^{2}})}^{19}}(3+2x)={{a}_{1}}+2{{a}_{2}}x+...+40{{a}_{40}}{{x}^{39}}\)

Cho \(x=1\) 

\(\begin{align}  & \Rightarrow 20{{(1+3.1+{{1}^{2}})}^{19}}(3+2.1)={{a}_{1}}+2{{a}_{2}}.1+...+40{{a}_{40}}{{.1}^{39}} \\ & \Rightarrow {{a}_{1}}+2{{a}_{2}}+...+40{{a}_{40}}={{20.5}^{20}} \\\end{align}\)

Vậy, \(S={{20.5}^{20}}\)

Chọn: D