Tìm số tự nhiên n thỏa mãn C0 n / 1.2 + C 1 n/ 2.3 + C2 n / 3.4 + ...+ Cn n / (n + 1)( n + 2) .

Cnkk+1k+2=n!k!n−k!k+1k+2=n+2!n−k!k+2!n+1n+2=Cn+2k+2n+1n+2.

Suy ra: ∑k=0nCnkk+1k+2=∑k=0nCn+2k+2n+1n+2

⇔Cn01.2+Cn12.3+Cn23.4+...+Cnnn+1n+2=Cn+22+Cn+23+Cn+24+...+Cn+2n+2n+1n+2 ∗.

Ta xét khai triển sau: 1+xn+2=Cn+20+x.Cn+21+x2.Cn+22+x3.Cn+23+...+xn+2.Cn+2n+2.

Chọn x=1→2n+2=Cn+20+Cn+21+Cn+22+Cn+23+...+Cn+2n+2.

Do đó: ∗⇔2100−n−3n+1n+2=2n+2−Cn+20−Cn+21n+1n+2⇔2100=2n+2⇔n=98.