### deepwork's blog

By deepwork, history, 12 months ago, ,

$a_1 = 1$
$a_2 = 5$
$a_n = a_{n-1} + n^2(a_{n-2}+1)$

Prove $a_n=(n+1)!-1$
P.S. Original problem

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 » 12 months ago, # |   0 Auto comment: topic has been updated by NewFlow (previous revision, new revision, compare).
 » 12 months ago, # |   +5 Putting $n = 1$ we get $a_1 = (1 + 1)! - 1 = 1$.Putting $n = 2$ we get $a_2 = (2 + 1)! - 1 = 6 - 1 = 5$.Now let the given statement is true for each $k < n$, let's prove this for $k = n$. $a_n = a_{n - 1} + n^2(a_{n - 2} + 1)$, then $a_n = n! - 1 + n^2((n - 1)! - 1 + 1) = n! - 1 + n^2 \cdot (n - 1)! = n! - 1 + n \cdot n! = n! \cdot (n + 1) - 1 = (n + 1)! - 1$. So, the statement is true for $k = n$.By the principle of mathematical induction this is true for each $n > 1$.
 » 12 months ago, # |   0 Auto comment: topic has been updated by NewFlow (previous revision, new revision, compare).