[SOLVED] How to prove reduction of this induction to formula?
$$$a_1 = 1$$$$$$a_2 = 5$$$$$$a_n = a_{n-1} + n^2(a_{n-2}+1)$$$
Prove $$$a_n=(n+1)!-1$$$P.S. Original problem
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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 $$$n > 1$$$.
CodeChef Starters 126 Solution Discussion
Auto comment: topic has been updated by prudent (previous revision, new revision, compare).
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 $$$n > 1$$$.
Auto comment: topic has been updated by prudent (previous revision, new revision, compare).