Giá trị của n thỏa mãn: 1c2n+1-2.22c2n+1+3.2^23c2n+1 bằng

Xét $k{C}_{2n+1}^k=k.\frac{\left(2n+1\right)!}{k!\left(2n+1-k\right)!}=\frac{\left(2n+1\right).\left(2n\right)!}{\left(k-1\right)!\left[2n-\left(k-1\right)\right]!}=\left(2n+1\right).C_{2n}^{k-1}$  với $1\le k\le 2n+1$ .

Ta có $VT=\left(2n+1\right)\left(C_{2n}^0-C_{2n}^1{2}^1+C_{2n}^2{2}^2-\mathrm{...}+C_{2n}^{2n}{2}^{2n}\right)=\left(2n+1\right){\left(1-2\right)}^{2n}=2n+1$  .

Do đó: $2n+1=2021\iff n=1010$ .