Chứng minh rằng:

$$\begin{gathered} a, {11}^{10}-1 chia hết 100 \\ b, {101}^{100}-1 chia hết 10 000 \\ c, \sqrt{10} \left[{\left(1+\sqrt{10}\right)}^{100}-{\left(1-\sqrt{10}\right)}^{100}\right] là một số nguyên \end{gathered}$$

Giải

 $$\begin{gathered} a, Ta có : {11}^{10}-1={\left(10+1\right)}^{10}-1=\sum _{k=0}^{10}{C}_{10}^k-1 \\ = 10.C_{10}^1-\sum _{k=2}^{10}{C}_{10}^k{10}^k chia hết cho 100 \\ Vì 10C_{10}^1=100 và \sum _{k=2}^{10}{C}_{10}^k{10}^k⋮100 \end{gathered}$$

b,

$$\begin{gathered} Ta có {100}^{100} -1 ={\left(100+1\right)}^{100}-1=\sum _{k=0}^{100}{C}_{100}^k{100}^k -1 \\ =1+C_{100}^1.100+\sum _{k=2}^{100}{C}_{100}^k{100}^k-1=1000+\sum _{k=2}^{100}{C}_{100}^k{100}^k⋮10000 \end{gathered}$$

c, 

$$\begin{gathered} Ta có \\ {\left(1+\sqrt{10}\right)}^{100}=\sum _{k=0}^{100}{C}_{100}^k{\left(\sqrt{10}\right)}^k \\ =\sum _{0\le k=2l\le 100}^{100}{C}_{100}^{2l}.10l+\sum _{0\le k=2l\le 100}^{100}{C}_{100}^{2l+1}.{10}^l.\sqrt{10} \\ =\sum _{l=0}^{50}{C}_{100}^{2l}.{10}^l+\sum _{l=0}^{49}{C}_{100}^{2l+1}.{10}^l.\sqrt{10} \left(1\right) \\ {\left(1-\sqrt{10}\right)}^{100}=\sum _{k=0}^{100}{C}_{100}^k{\left(-\sqrt{10}\right)}^k=\sum _{k=0}^{50}{C}_{100}^{2l}.{10}^l-\sum _{k=0}^{49}{C}_{100}^{2l+1}.{10}^l.\sqrt{10} \left(2\right) \\ Lấy \left(1\right) trừ \left(2\right) ta được \\ {\left(1+\sqrt{10}\right)}^{100}-{\left(1-\sqrt{10}\right)}^{100}=2\sum _{l=0}^{49}{C}_{100}^{2l+1}.{10}^l.\sqrt{10} \\ Suy ra \sqrt{10} \left[{\left(1+\sqrt{10}\right)}^{100}-{\left(1-\sqrt{10}\right)}^{100}\right]=2\sum _{l=0}^{49}{C}_{100}^{2l+1}{10}^{l+1} \in \mathrm{\mathbb{Z}} \end{gathered}$$