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

Xét:

M=C2n0+C2n1+C2n2+…+C2n2n−1+C2n2n;

N=C2n0−C2n1+C2n2−…−C2n2n−1+C2n2n;

P=C2n0+C2n2+C2n4+…+C2n2n−2+C2n2n;

Q=C2n1+C2n3+C2n5+…+C2n2n−3+C2n2n−1.

+) Ta có:

(x+1)2n=C2n0x2n+C2n1x2n−11+C2n2x2n−212+…+C2n2n−1x12n−1+C2n2n12n

=C2n0x2n+C2n1x2n−1+C2n2x2n−2+…+C2n2n−1x+C2n2n.

Cho x = 1, ta được:

(1+1)2n=C2n012n+C2n112n−1+C2n212n−2+…+C2n2n−11+C2n2n

=C2n0+C2n1+C2n2+…+C2n2n−1+C2n2n.

Vậy M=(1+1)2n=22n.

+) Ta có:

(x−1)2n=C2n0x2n−C2n1x2n−11+C2n2x2n−212−…−C2n2n−1x12n−1+C2n2n12n

=C2n0x2n−C2n1x2n−1+C2n2x2n−2−…−C2n2n−1x+C2n2n.

Cho x = 1, ta được:

(1−1)2n=C2n012n−C2n112n−1+C2n212n−2−…−C2n2n−11+C2n2n

=C2n0−C2n1+C2n2−…−C2n2n−1+C2n2n.

Vậy N=(1−1)2n=0

Ta có: P+Q=M=22n và P−Q=N=0 nên P=Q=22n:2=22n−1.

Áp dụng: C2n1+C2n3+…+C2n2n−1=2048⇒22n−1=2048⇒2n−1=11⇒n=6.