Giải bài 6 trang 58 – SGK môn Đại số và Giải tích lớp 11

Gợi ý:

a) Tách $11=10+1$ rồi khai triển $(10+1)^{10}$ theo công thức Niu -tơn.

b) Tách $101=100+1$ rồi khai triển $(100+1)^{100}$ theo công thức Niu -tơn.

c) Khai triển mỗi hạng tử trong ngoặc vuông rồi rút gọn.

a) Ta có:

$\begin{align} & {{11}^{10}}-1={{(10+1)}^{10}}-1=C_{10}^{0}+C_{10}^{1}10+...+C_{10}^{9}{{10}^{9}}+C_{10}^{10}{{10}^{10}}-1 \\ & =100+...+C_{10}^{9}{{10}^{9}}+C_{10}^{10}{{10}^{10}}\,\vdots\, 100 \\ \end{align}$

b) 

$\begin{align} & {{101}^{100}}-1={{(1+100)}^{100}}-1 \\ & =C_{100}^{0}+C_{100}^{1}100+C_{100}^{2}{{100}^{2}}+...+C_{100}^{99}{{100}^{99}}+C_{100}^{100}{{100}^{100}}-1 \\ & =C_{100}^{1}100+C_{100}^{2}{{100}^{2}}+...+C_{100}^{99}{{100}^{99}}+C_{100}^{100}{{100}^{100}} \\ & =10000+\sum\limits_{k=2}^{100}{C_{100}^{k}{{100}^{k}}}\,\,\vdots\, 10000 \\ \end{align}$

c) Ta có:

$\begin{align} & {{(1+\sqrt{10})}^{100}}=\sum\limits_{k=0}^{100}{C_{100}^{k}{{\left( \sqrt{10} \right)}^{k}}} \\ & =C_{100}^{0}+C_{100}^{1}{{\left( \sqrt{10} \right)}^{1}}+C_{100}^{2}{{\left( \sqrt{10} \right)}^{2}}+...+C_{100}^{99}{{\left( \sqrt{10} \right)}^{99}}+C_{100}^{100}{{\left( \sqrt{10} \right)}^{100}} \\ & {{(1-\sqrt{10})}^{100}}=\sum\limits_{k=0}^{100}{C_{100}^{k}{{\left( -\sqrt{10} \right)}^{k}}} \\ & =C_{100}^{0}-C_{100}^{1}{{\left( \sqrt{10} \right)}^{1}}+C_{100}^{2}{{\left( \sqrt{10} \right)}^{2}}-C_{100}^{3}{{\left( \sqrt{10} \right)}^{3}}+...+C_{100}^{98}{{\left( \sqrt{10} \right)}^{98}}-C_{100}^{99}{{\left( \sqrt{10} \right)}^{99}}+C_{100}^{100}{{\left( \sqrt{10} \right)}^{100}} \\ \end{align}$

Suy ra: 

$\begin{align} & {{(1+\sqrt{10})}^{100}}-{{(1-\sqrt{10})}^{100}}=\left[ C_{100}^{0}+C_{100}^{1}{{\left( \sqrt{10} \right)}^{1}}+C_{100}^{2}{{\left( \sqrt{10} \right)}^{2}}+...+C_{100}^{99}{{\left( \sqrt{10} \right)}^{99}}+C_{100}^{100}{{\left( \sqrt{10} \right)}^{100}} \right]\\&-\left[ C_{100}^{0}-C_{100}^{1}{{\left( \sqrt{10} \right)}^{1}}+C_{100}^{2}{{\left( \sqrt{10} \right)}^{2}}-C_{100}^{3}{{\left( \sqrt{10} \right)}^{3}}+...+C_{100}^{98}{{\left( \sqrt{10} \right)}^{98}}-C_{100}^{99}{{\left( \sqrt{10} \right)}^{99}}+C_{100}^{100}{{\left( \sqrt{10} \right)}^{100}} \right] \\ & =2\left[ C_{100}^{1}{{\left( \sqrt{10} \right)}^{1}}+C_{100}^{3}{{\left( \sqrt{10} \right)}^{3}}+...+C_{100}^{99}{{\left( \sqrt{10} \right)}^{99}} \right] \\ \end{align}$

$\begin{align} & \Rightarrow \sqrt{10}.[{{(1+\sqrt{10})}^{100}}-{{(1-\sqrt{10})}^{100}}]=2\left[ C_{100}^{1}{{\left( \sqrt{10} \right)}^{2}}+C_{100}^{3}{{\left( \sqrt{10} \right)}^{4}}+...+C_{100}^{99}{{\left( \sqrt{10} \right)}^{100}} \right] \\ & =2\left( C_{100}^{1}10+C_{100}^{3}{{10}^{2}}+...+C_{100}^{99}{{10}^{50}} \right)\in \mathbb{Z} \\ \end{align}$