Question

Biết rằng (2+x+2x³)¹⁵ = a₀ + a₁x + a₂x² + a₃x³ + ... + a₄₅x⁴⁵

Tính S= a₁ + a₂ + a₃ + .... + a₄₅

Answer 1

Đặt  \(A\left(x\right)=\left(2+x+2x^3\right)^{15}=a_0+a_1x+a_2x^2+a_3x^3+...+a_{45}x^{45}\)

Như vậy  \(A\left(0\right)=\left(2+0+2.0^3\right)^{15}=a_0+a_1.0+a_2.0^2+a_3.0^3+...+a_{45}.0^{45}=a_0\)

hay  \(a_0=\left(2+0+2.0^3\right)^{15}=2^{15}\)

Lại có  \(A\left(1\right)=\left(2+1+2.1^3\right)^{15}=a_0+a_1.1+a_2.1^2+a_3.1^3+...+a_{45}.1^{45}\)

                      \(=a_0+a_1+a_2+a_3+...+a_{45}=a_0+S=2^{15}+S\)

hay  \(2^{15}+S=\left(2+1+2.1^3\right)^{15}=5^{15}\)

\(\Rightarrow S=5^{15}-2^{15}\)