Cho khai triển

A

- Hướng dẫn giải

Đáp án A

Ta có: ${\left(1+x+x^2\right)}^n={\left[1+x\left(1+x\right)\right]}^n=\sum _{k=0}^n{C}_k^n{x}^k{\left(1+x\right)}^k$

$=\sum _{k=0}^n{C}_n^k{x}^k\left(\sum _{j=0}^k{C}_j^k{x}^k\right)\Rightarrow T_{k+1}=C_k^n{x}^k\left(\sum _{j=0}^k{C}_j^k{x}^k\right)$

Ta tính các số hạng như sau:

$T_0=1$;

$T_1=C_n^1{C}_n^{2}x+C_n^1{C}_1^1{x}^2=nx;T_2=C_n^2{C}_n^0{x}^2+C_n^2{C}_2^1{x}^3+C_n^2{C}_2^2{x}^4,\mathrm{....}$ 

Như vậy ta có:

$a_3=C_n^2{C}_2^1+C_n^3{C}_2^0;a_4=C_n^2{C}_2^2+C_n^3{C}_3^1+C_n^4{C}_4^0$   

Theo giả thiết  

$\frac{a_3}{14}=\frac{a_4}{41}\Rightarrow \frac{C_n^2{C}_2^1+C_n^3{C}_2^0}{14}=\frac{C_n^2{C}_2^2+C_n^3{C}_3^1+C_n^4{C}_4^0}{41}$

$\iff \frac{2.\frac{n\left(n-1\right)}{2!}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}}{14}=\frac{\frac{n\left(n-1\right)}{2!}+\frac{3n\left(n-1\right)\left(n-2\right)}{3!}+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}}{41}$

$\iff 21n^2-99n-1110=0\Rightarrow n=10$

Trong khai triển:

${\left(1+x+x^2\right)}^{10}=a_0+a_{1}x+a_2{x}^2+\mathrm{...}+a_{20}{x}^{20}$

cho x = 1 ta được: $S=a_0+a_1+a_2+\mathrm{...}+a_{20}=3^{10}$