Finding the Formula for nth Term of a Sequence: −32, −16, 112, 320, 530, ...

[kyle1]

Finding the Formula for nth Term of a Sequence: −32, −16, 112, 320, 530, ...

I just started calculus II and am already running into some trouble finding the formula for the nth term of this sequence:


{−3/2,−1/6,1/12,3/20,5/30...}

I have done some work of my own finding how much is added to the previous term to get the next term but I can't find any pattern that would lead me to a formula. I have completed numerous problems that are similar, just this sequence is giving me trouble.
Any advice helps! Thank you.

 

[mmm4444bot]

… finding the formula for the nth term of this sequence:

{−32, −16, 112, 320, 530...}

Please check to make sure that you have copied down these first five terms correctly.

 

[tkhunny]

Don't beat yourself up. There are ALWAYS infinitely many "next numbers". Always. Just pick one and prove it!

618 is a good next number, using constant 4th differences.

35205.4464864292 is a great next number, using constant 4th ratios.

-2, followed by -3,306 are two great choices, using constant 5th differences.

Seriously, these are just a couple ways to proceed. There are many, MANY more. Just pick one!

 

[stapel]

I just started calculus II and am already running into some trouble finding the formula for the nth term of this sequence:

{−32, −16, 112, 320, 530, ...}

I'm not seeing anything, but of course, with five points, you can always create a system of equations and create a quartic polynomial for these values. (You can see an example at the bottom of this page and another at the top of this page.)

I get this polynomial:

. . . . .\(\displaystyle p(n)\, =\, -\dfrac{23}{12}\, n^4\, +\, \dfrac{83}{6}\, n^3\, +\, \dfrac{251}{12}\, n^2\, -\, \dfrac{689}{6}\, n\, +\, 50\)

Whatever your sequence is meant to be, it doesn't appear to be anything that's "interesting", because the Online Encyclopedia of Integer Sequences (OEIS) doesn't have it: check it out.