$(1+\sqrt{2})^{3000}$ 的小数点后第

答案是 9。

利用 $(1+\sqrt2)^n+(1-\sqrt2)^n$ 为整数，且

$$0<(1-\sqrt2)^{3000}\ll 10^{-100}.$$

因此 $(1+\sqrt2)^{3000}$ 比某个整数略小一个极其微小的正数，小数点后很长一段全为 9，第 100 位也是 9。

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Original Explanation

By the Binomial Theorem and the fact that $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer, note also

$$0 < (1 - \sqrt{2})^{3000} \ll 10^{-100}.$$

Thus, $(1 + \sqrt{2})^{3000}$ is extremely close to an integer but slightly less than that integer by about $(1 - \sqrt{2})^{3000}$. Therefore, the 100th digit to the right of the decimal point is 9.