Find a polynomial function f(n) that follows the following sequence: 1,1,2,4,7,11,16,21

Question

my question is as follows:

Find a polynomial function $f(n)$ such that $f(1),f(2),…,f(8)$ is exactly the following squence: 1,1,2,4,7,11,16,22.

(Hint: how does the sum $\sum_{n=2}^{i=0}i$ come into this?)

I know that the sequence increases by adding an integer that increases by 1 each time, but I have no idea how to express this in a function. I'm also not quite sure what the hint means either? How do I approach/solve this?

Answer 1

Hint:

You have the sequence $$1,1,2,4,7,11,16,22,\dots\tag{1}$$ Now look at the difference of consecutive elements ($a_{n+1}-a_n$): $$0,1,2,3,4,5,6,\dots\tag{2}$$ Look again at the difference of elements in $(2)$: $$1,1,1,1,1,1,\dots\tag{3}$$ Since the sequence becomes constant after two computing the difference twice, what degree must $f(x)$ have?

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Now you know that $f(n)=an^2+bn+c$, and can find the coefficients.

• $f(1)=1\Rightarrow a+b+c=1$
• $f(2)=1\Rightarrow 4a+2b+c=1$
• $f(3)=2\Rightarrow 9a+3b+c=2$

This will give $$f(n)=\frac{n^2}{2}-\frac{3}{2}n+2$$

Answer 2

$f(n)=\frac{n^2}{2}-\frac{3n}{2}+2$

For you know it is quadratic (see the other answer). If the second difference is 1 (as it is in this case) then half is the coefficient. See where it would cut the y-axis by going back one place to $n=0$ and then work out the $n$ coefficient by trial and error.

Answer 3

A polynomial of degree n-1 has n coefficients so those coefficients can be chosen to fit n conditions. Here, you have 8 conditions: f(1)= 1, f(2)= 1, f(3)= 2, f(4)=4, f(5)= 7, f(6)= 11, f(6)= 16, and f(8)= 22 so we can fit a unique polynomial of degree 7 to that. Write $f(x)= a_0+ a_1x+ a_2x^2+ a_3x^3+ a_4x^4+ a_5x^5+ a_6x^6+ a_7x^7$ and we must have $f(1)= a_0+ a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7= 1$ $f(2)= a_0+ 2a_1+ 4a_2+ 8a_3+ 16a_4+ 32a_5+ 64x^6+ 128a_7= 1$ $f(3)= a_0+ 3a_1+ 9a_2+ 27a_3+ 81a_4+ 243a_5+ 729a_6+ 2187a_7= 2$ $f(4)= a_0+ 4_1+ 16a_2+ 64a_3+ 256a_4+ 1024a_5+ 4096a_6+ 16384a_7= 4$ $f(5)= a_0+ 5a_1+ 25a_2+ 125a_3+ 625a_4+ 3125a_5+ 15625a_6+ 78125a_7= 7$ $f(6)= a_0+ 6a_1+ 36a_2+ 215a_3+ 1296a_4+ 7776a_5+ 46656a_6+ 279936a_7= 11$ $f(7)= a_0+ 7a_1+ 49a_2+ 343a_3+ 2401a_4+ 16807a_5+ 117649a_6+ 823543a_7= 16$

Solve those 8 linear equations for the 8 coefficients.