Chứng minh rằng Áp dụng: Tìm số nguyên dương n thoả mãn

Xét:

$M=C_{2n}^0+C_{2n}^1+C_{2n}^2+\dots +C_{2n}^{2n-1}+C_{2n}^{2n}$;

$N=C_{2n}^0-C_{2n}^1+C_{2n}^2-\dots -C_{2n}^{2n-1}+C_{2n}^{2n}$;

$P={\text{C}}_{2n}^0+{\text{C}}_{2n}^2+{\text{C}}_{2n}^4+\dots +{\text{C}}_{2n}^{2n-2}+{\text{C}}_{2n}^{2n}$;

$Q=C_{2n}^1+C_{2n}^3+C_{2n}^5+\dots +C_{2n}^{2n-3}+C_{2n}^{2n-1}.$

+) Ta có:

$=C_{2n}^0{x}^{2n}+C_{2n}^1{x}^{2n-1}+C_{2n}^2{x}^{2n-2}+\dots +C_{2n}^{2n-1}x+C_{2n}^{2n}.$

Cho x = 1, ta được:

$={\text{C}}_{2n}^0+{\text{C}}_{2n}^1+{\text{C}}_{2n}^2+\dots +{\text{C}}_{2n}^{2n-1}+{\text{C}}_{2n}^{2n}.$

Vậy $M={(1+1)}^{2n}=2^{2n}$.

+) Ta có:

$={\text{C}}_{2n}^0{x}^{2n}-{\text{C}}_{2n}^1{x}^{2n-1}+{\text{C}}_{2n}^2{x}^{2n-2}-\dots -{\text{C}}_{2n}^{2n-1}x+{\text{C}}_{2n}^{2n}.$

Cho x = 1, ta được:

${(1-1)}^{2n}={\text{C}}_{2n}^0{1}^{2n}-{\text{C}}_{2n}^1{1}^{2n-1}+{\text{C}}_{2n}^2{1}^{2n-2}-\dots -{\text{C}}_{2n}^{2n-1}1+{\text{C}}_{2n}^{2n}$

$={\text{C}}_{2n}^0-{\text{C}}_{2n}^1+{\text{C}}_{2n}^2-\dots -{\text{C}}_{2n}^{2n-1}+{\text{C}}_{2n}^{2n}.$

Vậy $N={(1-1)}^{2n}=0$

Ta có: $P+Q=M=2^{2n}$ và $P-Q=N=0$ nên $P=Q=2^{2n}:2=2^{2n-1}$.

Áp dụng: $C_{2n}^1+C_{2n}^3+\dots +C_{2n}^{2n-1}=2048\Rightarrow 2^{2n-1}=2048\Rightarrow 2n-1=11\Rightarrow n=6.$