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the sum of the first $n$ terms of a certain sequence is $n(n 1)(n 2).$ find the tenth term of the sequence.

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Answer: Let $a_n$ denote the $n$th term of the sequence. We know that the sum of the first $n$ terms of the sequence is $n(n-1)(n-2)$, so we can set up the following equation:

$$a_1 + a_2 + \cdots + a_n = n(n-1)(n-2)$$

Since we want to find the tenth term of the sequence, we can substitute $n = 10$ into the above equation to get:

$$a_1 + a_2 + \cdots + a_{10} = 10(10-1)(10-2)$$

Simplifying the right-hand side gives:

$$a_1 + a_2 + \cdots + a_{10} = 720$$

We know that the sum of the first nine terms of the sequence is:

$$a_1 + a_2 + \cdots + a_9 = 9(9-1)(9-2) = 504$$

Subtracting this equation from the previous one, we get:

$$a_{10} = (a_1 + a_2 + \cdots + a_{10}) - (a_1 + a_2 + \cdots + a_9) = 720 - 504 = 216$$

Therefore, the tenth term of the sequence is 216.

Explanation:

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