Find the number for the question mark

Question

What is the number for the question mark? (My original puzzle)

3 $\rightarrow$ 39

1 $\rightarrow$ 3

4 $\rightarrow$ 84

5 $\rightarrow$ 155

2 $\rightarrow$ 14

10 $\rightarrow\ \large?$

Answer 1

By noting that each number maps to a multiple of itself:
$3\rightarrow3*13$
$1\rightarrow1*3$
$4\rightarrow4*21$
$5\rightarrow5*31$
$2\rightarrow2*7$
And that the quotient increases quadratically
$1:3$
$2:7=3+4$
$3:13=7+6$
$4:21=13+8$
$5:31=21+10$
The mapping must be a cubic function.
The above sequence is equal to $n^2+n+1$, where n is the input.
The original sequence is multiplied by $n$ again. So the function can be described by $n^3+n^2+n$.
So the final answer is $10^3+10^2+10=1110$.

Answer 2

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)
2 x 7 = 14 (7 is the 4th positive odd number)
3 x 13 = 39 (13 is the 7th positive odd number)
4 x 21 = 84 (21 is the 11th positive odd number)
5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

Answer 3

I came to the same conclusion as the others through a different method maybe, sorry if this is exactly what someone else said or I use some funky syntax:

I noted that

3/1 = 3
14/2 = 7
39/3 = 13
84/4 = 21
155/5 = 31
The difference between the RHS of each group increases by 2 each time.

Therefore:

Extrapolating the differences (up to 10) we get:
12, 14, 16, 18 and 20 Adding the differences to the RHs of the first set of equations starting from the last one (31), results in 111

The answer should be:

1,110, since 111 * 10 = 1,110

I feel like I probably didn't do a great job of explaining this and if that's the case, apologies but I'm new to this community and I'm not very familiar with how questions should be answered.