How to get the general term for a quartic sequence really need help

Question

Is it possible to find the general term for a quartic sequence and if so how do you do it? The sequence I am using is 1,9,36,100,225,441, 784, 1296, 2025, 3025 I am only interested in finding the general term in the form ax^4 + bx^3 + cx^2.... Even if you can`t give an answer could you please tell me if it's even possible Thank You

Answer 1

From the difference table (mentioned by @lhf in a comment), the leftmost column implies formula $$1\binom{x}{0}+8\binom{x}{1}+19\binom{x}{2}+18\binom{x}{3}+6\binom{x}{4},$$ which simplifies to $$\frac{1}{4}x^4 + \frac{3}{2} x^3 + \frac{13}{4} x^2 + 3 x + 1.$$ If your first sequence element corresponds to $x=1$ instead of $x=0$, then shift the formula: $$\frac{1}{4}(x-1)^4 + \frac{3}{2} (x-1)^3 + \frac{13}{4} (x-1)^2 + 3 (x-1) + 1 = \color{blue}{\frac{1}{4}x^4 + \frac{1}{2}x^3 + \frac{1}{4}x^2} = \left(\frac{x(x+1)}{2}\right)^2.$$

Answer 2

Say f(x)=ax⁴+bx³+cx²+dx+e. This is the general form of a quartic polynomial. From the information you've given, f(1)=1, f(2)=9, f(3)=36, f(4)=100, and f(5)=225. Substitute in these values instead of x and you'll have five equations and five unknowns which should be pretty easy to solve. Then, using the values you get for a,b,c,d, and e, you can get the formula for the nth term of the sequence.

Answer 3

Just looking at this, the terms are

$1^2, 3^2, 6^2, 10^2, 15^2, 21^2, 28^2, 36^2, 45^2, 55^2 $ which are the squares of the triangular numbers or $\left(\dfrac{n(n+1)}{2}\right)^2 =\dfrac{n^2(n^2+2n+1)}{4} =\dfrac{n^4+2n^3+n^2}{4} $.