$$$a_1 = 1$$$
$$$a_2 = 5$$$
$$$a_n = a_{n-1} + n^2(a_{n-2}+1)$$$
Prove $$$a_n=(n+1)!-1$$$
P.S. Original problem
# | User | Rating |
---|---|---|
1 | jiangly | 3640 |
2 | Benq | 3593 |
3 | tourist | 3572 |
4 | orzdevinwang | 3561 |
5 | cnnfls_csy | 3539 |
6 | ecnerwala | 3534 |
7 | Radewoosh | 3532 |
8 | gyh20 | 3447 |
9 | Rebelz | 3409 |
10 | Geothermal | 3408 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | adamant | 164 |
3 | awoo | 162 |
4 | TheScrasse | 160 |
5 | nor | 159 |
6 | maroonrk | 156 |
7 | SecondThread | 154 |
8 | pajenegod | 147 |
9 | Um_nik | 145 |
9 | BledDest | 145 |
$$$a_1 = 1$$$
$$$a_2 = 5$$$
$$$a_n = a_{n-1} + n^2(a_{n-2}+1)$$$
Prove $$$a_n=(n+1)!-1$$$
P.S. Original problem
Name |
---|
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).