Use polynomial fitting to find the formula for the nth term of the sequence which starts, (an)n>=1

Question

11,45,119,245,435,… Note the first term above is a1, not a0.

Answer 1

Create a table:

1 11

2 45

3 119

4 245

5 435

From here, you can go 2 ways. Method 1 is to put the table into your TI-84 as lists L1 (the 1, 2, 3, 4, 5) and L2 (the 11,45,119,245,435) and then use "stat" then "CALC", then, starting with "5:Quadreg" calculate the associated coefficients. Ensure the diagnostics is turned on so you get the R² value. If you get R² = 1, this is the correct polynomial curve fit. If not, run "6:CubicReg" and, if necessary, run "7:QuartReg". Again you are looking for R² = 1 (a perfect fit). Using the table above, I got an R2 = 1 on "6:CubicReg". The calculator will give you the polynomial coefficients to use.

Method 2, Using the Table, calculate the 1st difference, the 2nd difference, etc until you find a difference that is constant. This will be the degree of the polynomial. Example:

# Value 1st Diff 2nd Diff 3rd Diff

1 11

2 45 34

3 119 74 40

4 245 126 52 12

5 435 190 64 12

Since the 3rd difference is constant, the "perfect fit" polynomial is a 3rd degree polynomial meaning it fits the pattern:

f(x) = ax³ + bx² + cx + d

Now use the data points to solve for a, b, c, and d:

The first term of 11, means f(1) = 11 or:

11 = a(1)³ + b(1)² + c(1) + d

a + b + c + d = 11

The 2nd term of 45, means f(2) = 45 or:

45 = a(2)³ + b(2)² + c(2) + d

8a + 4b + 2c + d = 45

The 3rd term of 119, means f(3) = 119 or:

119 = a(3)³ + b(3)² + c(3) + d

27a + 9b + 3c + d = 119

The 4th term of 245, means f(4) = 245 or:

245 = a(4)³ + b(4)² + c(4) + d

64a + 16b + 4c + d = 245

Now, solve the 4 equations in 4 unknown you just developed

1) a + b + c + d = 11

2) 8a + 4b + 2c + d = 45

3) 27a + 9b + 3c + d = 119

4) 64a + 16b + 4c + d = 245

Again, you can use your calculator to solve these. Otherwise, use elimination to do so.