2005 AIME II 第 7 题

Question

设

\begin{aligned} x &= \scriptsize \frac{4}{(\sqrt{5} + 1)(\sqrt[4]{5} + 1)(\sqrt[8]{5} + 1)(\sqrt[16]{5} + 1)} \end{aligned}\text{。}

求

(x + 1)^{48}

。

Let

\begin{aligned} x &= \scriptsize \frac{4}{(\sqrt{5} + 1)(\sqrt[4]{5} + 1)(\sqrt[8]{5} + 1)(\sqrt[16]{5} + 1)}. \end{aligned}

Find

(x + 1)^{48}.

答案

令

y = \sqrt[16]{5}

，并将分子和分母同乘以

y - 1

。

Let

y = \sqrt[16]{5}

and multiply the numerator and denominator by

y - 1

反复使用平方差公式会把分母化简为

y^{16} - 1 = 4

。

Repeated difference of squares collapses the denominator to

y^{16} - 1 = 4

令

y = \sqrt[16]{5}

。分子和分母同乘以

y - 1

，反复使用平方差公式后，分母中的各项会逐次相消：

\begin{aligned} x &= \scriptsize \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} \\ &= \frac{4(y-1)}{y^{16} - 1} \\ &= \frac{4(y-1)}{4} \\ &= y - 1 \end{aligned}\text{。}

因此

x + 1 = y = 5^{\frac{1}{16}}

，且

(x+1)^{48} = 5^{\frac{48}{16}} = 5^3 = 125

。

Let

y = \sqrt[16]{5}.

Multiplying the numerator and denominator by

y - 1

telescopes the denominator by repeated difference of squares:

\begin{aligned} x &= \scriptsize \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} \\ &= \frac{4(y-1)}{y^{16} - 1} \\ &= \frac{4(y-1)}{4} \\ &= y - 1. \end{aligned}

Hence

x + 1 = y = 5^{\frac{1}{16}},

and

(x+1)^{48} = 5^{\frac{48}{16}} = 5^3 = 125.

其他年份的第 7 题