Question

[Phạm Thu Hương]

Cho đa thức f(x)=\(\left(x^2+x+2\right)^{20}\)=\(a_0x^{40}+a_1x^{39}+a_2x^{38}+...+a_{39}x+a_{40}\)

Tính \(S=a_0+a_1+a_2+...+a_{39}+a_{40}\)

Answer 1

[Hồ Thu Giang]

Thay x = 1 

=> f(1) = \(\left(1^2+1+2\right)^{20}\)= \(a_0.1^{40}+a_1.1^{39}+a_2.1^{38}+...+a_{39}.1+a_{40}\)

= \(a_0+a_1+a_2+...+a_{39}+a_{40}\)= S

=> S = \(\left(1^2+1+2\right)^{20}\)

=> S = \(4^{20}\)