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Algebra Difficulty 3.7 AMC 10/12 Find the answer China

Sequence {an}\{a_n\} satisfies a1=2a_1 = 2 and an+1=(n+1)an−na_{n+1} = (n+1)a_n - n, n=1,2,…n = 1, 2, \dots. Then the general term formula of {an}\{a_n\} is ______.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By the condition, we have an+1−1=(n+1)(an−1)a_{n+1} - 1 = (n+1)(a_n - 1). Therefore,
an−1=n(an−1−1)=n(n−1)(an−2−1)=…=n(n−1)…2(a1−1)=n!, \begin{aligned} a_n - 1 &= n(a_{n-1} - 1) = n(n-1)(a_{n-2} - 1) = \dots \\ &= n(n-1)\dots2(a_1 - 1) = n!, \end{aligned}
namely, an=n!+1a_n = n! + 1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.