有趣的证明题

证明：\(\dfrac{10^{2n} - 1}{9} - \dfrac{2(10^n - 1)}{9} = (\dfrac{10^n - 1}{3})^2\)

第一种证法（直接推）：

\[\begin{align} \text{LHS}(左式) & = \dfrac{10^{2n} - 1 - 2(10^n - 1)}{9} \\ & = \dfrac{10^{2n} - 1 - 2 \cdot 10^n + 2}{9} \\ & = \dfrac{(10^{n})^2 - 2 \cdot 10^n + 1}{9} \\ & = \dfrac{(10^n - 1)^2}{9} \\ & = (\dfrac{10^n - 1}{3})^2 = \text{RHS}(右式) \\ \end{align} \]

\[\therefore \dfrac{10^{2n} - 1}{9} - \dfrac{2(10^n - 1)}{9} = (\dfrac{10^n - 1}{3})^2 \]

第二种证法（数学归纳法）：

\[\begin{array}{l} 当\ n = 1\ 时，\text{LHS} = 11 - 2 = 9 = 3^2 = \text{RHS}\ \therefore \ 成立 \\ 当\ n = k\ (k\ 为正整数且\ k > 1)\ 时，令\ \dfrac{10^{2n} - 1}{9} - \dfrac{2(10^n - 1)}{9} = (\dfrac{10^n - 1}{3})^2 \\ 当\ n = k + 1\ 时: \end{array} \]

\[\begin{align} \text{LHS} & = \dfrac{10^{2k + 2} - 1}{9} - \dfrac{2 \cdot 10^{k + 1} - 2}{9} \\ & = \dfrac{100 \cdot 10^{2k} - 1}{9} - \dfrac{20 \cdot 10^k - 2}{9} \\ & = \dfrac{10^{2k} + 99 \cdot 10^{2k} - 1}{9} - \dfrac{2 \cdot 10^k + 18 \cdot 10^k - 2}{9} \\ & = \dfrac{10^{2k} - 1}{9} - \dfrac{2 \cdot 10^k - 2}{9} + 11 \cdot 10^{2k} - 2 \cdot 10^k \\ & = [\dfrac{10^{2k} - 1}{9} - \dfrac{2 \cdot (10^k - 1)}{9}] + 11 \cdot 10^{2k} - 2 \cdot 10^k \\ & = (\dfrac{10^k - 1}{3})^2 + 11 \cdot 10^{2k} - 2\cdot 10^k \\ & = \dfrac{10^{2k} - 2 \cdot 10^k + 1}{9} + 11 \cdot 10^{2k} - 2 \cdot 10^k \\ & = \dfrac{10^{2k} - 2 \cdot 10^k + 1 + 99 \cdot 10^{2k} - 18 \cdot 10^k}{9} \\ & = \dfrac{100 \cdot 10^{2k} - 20 \cdot 10^k + 1}{9} \\ & = \dfrac{10^{2k + 2} - 2 \cdot 10^{k + 1} + 1}{9} \\ & = \dfrac{(10^{k + 1})^2 - 2 \cdot 10^{k + 1} + 1}{9} \\ & = \dfrac{(10^{k + 1} - 1)^2}{9} \\ & = (\dfrac{10^{k + 1} - 1}{3})^2 \\ & = (\dfrac{10^n - 1}{3}) = \text{RHS} \ \therefore 成立 \end{align} \]

\[\therefore \dfrac{10^{2n} - 1}{9} - \dfrac{2(10^n - 1)}{9} = (\dfrac{10^n - 1}{3})^2 \]