How to find the nth term of this sequence?

You need to check what type of equation replicates the data given.

Clearly a 1st order (linear) equation wont work.

If you next try a 2nd order solution: y = ax² + bx + c

it also wont work (not shown).

A 3rd order solution will however work perfectly: y = ax³ + bx² + cx + d

The way to get the 3rd order coefficients: a, b, c, d is to solve four simultaneous

equations.

    y = ax³ + bx² + cx + d

   31 = a + b + c + d ... substituted the 1st data point (1, 31)

   41 = 8a + 4b + 2c + d ... substituted the 2nd data point (2, 41)

   48 = 27a + 9b + 3c + d ... substituted the 3d data point (3, 48)

   59 = 64a + 16b + 4c + d ... substituted the 3d data point (4, 59)

  ▬▬▬▬▬▬▬▬▬▬▬▬▬<solve simultaneously>

 Solution:  a = 7 ⁄ 6  b = -17 ⁄ 2  c = 82 ⁄ 3  d = 11

... it's helpful to have a calculator that can do this for you.

... I had a Ti calculator that could solve nine equations with nine variables !

... It took the thing more than ten minutes to get the nine variables. LOL

... Solving nine equations by hand would take me about a year.

To disguise of the fractional coefficients a, b, c, and d, multiply them by 6

and include a (⅙) multiplier to cancel it out.

 6 multiplied by:   { a = 7 ⁄ 6  b = -17 ⁄ 2  c = 82 ⁄ 3  d = 11 }

     equals { a = 7  b = -  51  c = 164  d = 66 }

 ... replacing "x" with "n" and "y" with the functional symbol : a { n }

 a { n } = (⅙) • (7n³ – 51n² + 164n + 66)  for n = 1, 2, 3, ...

 a { n } = 31, 41, 48, 59, 81, 121, 186, 283, 419, 601

The next number is 67, I think.

Because 31 + 17 gives you 48 (3rd term), and 41 + 18 gives you 59 (4th term), I added 19 to 48.