Tìm số tự nhiên n thỏa mãn $\frac{C_n^0}{1.2}+\frac{C_n^1}{2.3}+\frac{C_n^2}{3.4}+\mathrm{...}+\frac{C_n^n}{\left(n+1\right)\left(n+2\right)}=\frac{2^{100}-n-3}{\left(n+1\right)\left(n+2\right)}$.

$\frac{C_n^k}{\left(k+1\right)\left(k+2\right)}=\frac{n!}{k!\left(n-k\right)!\left(k+1\right)\left(k+2\right)}=\frac{\left(n+2\right)!}{\left(n-k\right)!\left(k+2\right)!\left(n+1\right)\left(n+2\right)}=\frac{C_{n+2}^{k+2}}{\left(n+1\right)\left(n+2\right)}$.

Suy ra: $\sum _{k=0}^n\frac{C_n^k}{\left(k+1\right)\left(k+2\right)}=\sum _{k=0}^n\frac{C_{n+2}^{k+2}}{\left(n+1\right)\left(n+2\right)}$

$\iff \frac{C_n^0}{1.2}+\frac{C_n^1}{2.3}+\frac{C_n^2}{3.4}+\mathrm{...}+\frac{C_n^n}{\left(n+1\right)\left(n+2\right)}=\frac{C_{n+2}^2+C_{n+2}^3+C_{n+2}^4+\mathrm{...}+C_{n+2}^{n+2}}{\left(n+1\right)\left(n+2\right)}$ $\left(\ast \right)$.

Ta xét khai triển sau: ${\left(1+x\right)}^{n+2}=C_{n+2}^0+x.C_{n+2}^1+x^2.C_{n+2}^2+x^3.C_{n+2}^3+\mathrm{...}+x^{n+2}.C_{n+2}^{n+2}$.

Chọn $x=1\stackrel{}{\to }{2}^{n+2}=C_{n+2}^0+C_{n+2}^1+C_{n+2}^2+C_{n+2}^3+\mathrm{...}+C_{n+2}^{n+2}$.

Do đó: $\left(\ast \right)\iff \frac{2^{100}-n-3}{\left(n+1\right)\left(n+2\right)}=\frac{2^{n+2}-C_{n+2}^0-C_{n+2}^1}{\left(n+1\right)\left(n+2\right)}\iff 2^{100}=2^{n+2}\iff n=98$.