数列 $T_1, T_2, T_3, T_4,\dots$ 的前四项如下： $T_1 = 2^2 - 1 = 3$ $T_2 = 3^2 - 2 = 7$ $T_3 = 4^2 - 3 = 13$ $T_4 = 5^2 - 4 = 21$

(a)
求该数列的第 7 项 $T_7$ 。

答案 $T_7 = \dots\dots\dots\dots\dots\dots\dots\dots\dots$
(b)
写出该数列第 $n$ 项的表达式。

答案 $T_n = \dots\dots\dots\dots\dots\dots\dots\dots\dots$
(c)
Kelvin 声称，当 $n$ 是正整数时，相邻两项之差 $T_{n+1}-T_n$ 总是偶数。你同意吗？请写出清晰的计算/推导过程来解释你的答案。

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

1. Extend the given pattern to the 7th term and calculate it.

From the pattern,
$T_7=8^2-7=64-7=57.$

答案

$T_7=57$

1. Generalise the given pattern to the $n$th term.

From the pattern,

$T_1=2^2-1=(1+1)^2-1$

$T_2=3^2-2=(2+1)^2-2$

$T_3=4^2-3=(3+1)^2-3$

So, replacing the term number by $n$ ,

$T_n=(n+1)^2-n$ .

答案

$T_n=(n+1)^2-n$

1. Write expressions for two consecutive terms using the nth term.

From part (b),
$T_n=(n+1)^2-n$ .

So
$T_{n+1}=((n+1)+1)^2-(n+1)=(n+2)^2-(n+1).$

2. Find and simplify the difference between consecutive terms.

$T_{n+1}-T_n=[(n+2)^2-(n+1)]-[(n+1)^2-n]$
$=n^2+4n+4-n-1-(n^2+2n+1-n)$
$=n^2+3n+3-(n^2+n+1)$
$=2n+2.$

3. Factorise the result and explain why it is always even.

$2n+2=2(n+1).$
Since $n$ is a positive integer, $n+1$ is an integer. Therefore $2(n+1)$ is a multiple of 2, so it is always even.

答案

Yes, I agree. Since $T_{n+1}-T_n=2n+2=2(n+1)$ , it is always even for every positive integer $n$ .

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