题目

Problem

Three consecutive terms in a sequence of real numbers are given by

k,\ 1+2k \text{ and } 3+3k

where $k$ is a constant.

Use proof by contradiction to show that this sequence is not a geometric sequence.

(5)

题目中文翻译

某个实数数列中连续三项分别为

k,\ 1+2k \text{ 和 } 3+3k

其中 $k$ 为常数。

用反证法证明这个数列不是等比数列。

(5)

解答

展开

假设这三项构成等比数列。等比数列连续三项满足“中项的平方等于前后两项的乘积”，由此得到关于实数 $k$ 的二次方程。配方后会发现左边恒正，不可能等于零，从而产生矛盾。

展开

Suppose, for a contradiction, that the three terms form a geometric sequence.

For three consecutive terms of a geometric sequence, the square of the middle term equals the product of the other two. Therefore,

(1+2k)^2=k(3+3k).

Expanding and simplifying gives

\begin{align*} 1+4k+4k^2=&\,3k+3k^2\\ k^2+k+1=&\,0. \end{align*}

However,

\begin{align*} &\,k^2+k+1\\ =&\,\bigg(k+\frac12\bigg)^2+\frac34\\ >&\,0 \end{align*}

for every real value of $k$ . Thus $k^2+k+1$ cannot equal zero, which contradicts the assumption that the terms form a geometric sequence.

Therefore, the sequence is not geometric.

展开

仍假设三项构成等比数列，并用中项平方等于两端乘积得到 $k^2+k+1=0$ 。这次不配方，而是计算判别式；判别式为负，方程没有实数根，与 $k$ 为实数矛盾。

展开

Suppose, for a contradiction, that the sequence is geometric. Then the three consecutive terms satisfy

(1+2k)^2=k(3+3k).

Thus

\begin{align*} 1+4k+4k^2=&\,3k+3k^2\\ k^2+k+1=&\,0. \end{align*}

The discriminant of this quadratic equation is

\begin{align*} b^2-4ac=&\,1^2-4(1)(1)\\ =&\,-3<0. \end{align*}

Hence the equation has no real roots. This contradicts $k$ being real, so the assumption is false. Therefore, the sequence is not geometric.