$u_1=-2,$ $u_2=-1,$ $u_{n+1}=\sqrt[3]{n(u_n^{2}+1)+2u_{n-1}}$ Find $u_{2021}$ Let $(u_n)$ sequence satisfy: $u_1=-2, u_2=-1:$ $u_{n+1}=\sqrt[3]{n(u_n^{2}+1)+2u_{n-1}}.$
Find $u_{2021}.$
I tried to find $u_n=f(n)$ but seem so difficult. Can anyone help me with this problem? Thank you
 A: Hint:
First, manually check $u_3, u_4, u_5, u_6, u_7$.
Then, with the hypothesis that $u_n = n-3$,
consider
$$u_{n+1} = \sqrt[3]{\{ ~n \times ~[(n-3)^2 + 1] ~\} + 2(n - 4)}.\tag1 $$
In (1) above, just multiply everything out, under the radical sign.  Then, since you are (in effect) trying to prove that $u_{n+1} = (n-2)$, compare the expression under the radical sign with $(n-2)^3.$
A: For the calculations we can use the symbolic calculation software. Using Mathematica for example with the function RecurrenceTable as follows:

*

*RecurrenceTable[{u[n + 1] == (n*(u[n]^2 + 1) + 2*u[n - 1])^(1/3),  u[1] == -2, u[2] == -1},  u, {n, 20}]
Then we obtain

*

*{-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17}
Then it is reasonable to assume inductively that the closed formula for $(u_{n})_{n\in \mathbb{N}}$ is given by $u_{n}=n-3$.
Mathemematical Induction on $n\in \mathbb{N}$.
Define the stament for all natural number $n$
$$P(n): \quad u_{n}=n-3$$
Basis case:

*

*It's clear using the calculations above.

Induction step:

*

*Suppose that for any $k\geqslant 1$, $P(k)$ holds, then we need to show that $P(k+1)$ also holds. Algebraically, we can see:

$$u_{k+1}=\sqrt[3]{k(u_{k}^{2}+1)+2u_{k-1}} \overset{u_{k}=k-3}{\underset{ u_{k-1}=k-4}{\implies}}=\sqrt[3]{k((k-3)^{2}+1)+2(k-4)}\overset{\text{expand}}{=}\sqrt[3]{(k-2)^{3}}=(k+1)-3. $$
That is the statement $P(k+1)$ is also true.
Therefore $u_{n}=n-3$ for all $n\in \mathbb{N}$.
Then $u_{2021}=2021-3=2018$.
