某数列的前四项如下所示。 $T_1 = 2^2 + 7$ $T_2 = 3^2 + 12$ $T_3 = 4^2 + 17$ $T_4 = 5^2 + 22$ 解释为什么对于所有的 $n$ 值，$T_n$ 的值必须是奇数。 答案 ... 某数列前 $n$ 项的乘积由 $2n^2 + 3n$ 给出。 求该数列的第 12 项。 答案 ...

1. Write the general expression for the nth term from the given pattern.

The squared numbers are $2^2,3^2,4^2,5^2$, so the squared part is $(n+1)^2$.\n\nThe added numbers are $7,12,17,22$, increasing by $5$, so the added part is $5n+2$.\n\nHence\n$T_n=(n+1)^2+(5n+2).$

2. Explain why the two parts always have opposite parity, so their sum is odd.

If $n$ is odd, then $n+1$ is even, so $(n+1)^2$ is even. Also, $5n$ is odd, so $5n+2$ is odd.\n\nIf $n$ is even, then $n+1$ is odd, so $(n+1)^2$ is odd. Also, $5n$ is even, so $5n+2$ is even.\n\nTherefore $T_n$ is always an odd number plus an even number, which is odd.

1. Find the product of the first 12 terms and the product of the first 11 terms.

Let $P_n=2n^2+3n$.

$P_{12}=2(12)^2+3(12)=288+36=324$

$P_{11}=2(11)^2+3(11)=242+33=275$

2. Divide consecutive products to find the 12th term.

Since

$P_{12}=T_1T_2\cdots T_{11}T_{12}$

and

$P_{11}=T_1T_2\cdots T_{11},$

$T_{12}=\frac{P_{12}}{P_{11}}=\frac{324}{275}.$