Answer:
[tex]\huge\boxed{a=\dfrac{3}{2};\ b=\dfrac{13}{3}}[/tex]
Step-by-step explanation:
We have the formula:
[tex]f(n)=\dfrac{1}{6}n^3+an^2+bn[/tex]
We know
for n = 1, f(n) = 6
for n = 2, f(n) = 16.
Substitute:
[tex]\dfrac{1}{6}(1^3)+a(1^2)+b(1)=6\\\\\dfrac{1}{6}+a+b=6\qquad(1)\\---------------------\\\dfrac{1}{6}(2^3)+a(2^2)+b(2)=16\\\\\dfrac{1}{6}(8)+4a+2b=16\\\\\dfrac{8}{6}+4a+2b=16\qquad(2)[/tex]
We have the system of equations.
Multiply both sides of the equation (1) by (-2):
[tex](-2)\left(\dfrac{1}{6}\right)+(-2)a+(-2)b=(-2)(6)\\\\-\dfrac{2}{6}-2a-2b=-12[/tex]
Add both sides of the equations:
[tex]\underline{+\left\{\begin{array}{ccc}-\frac{2}{6}-2a-2b=-12\\\\\frac{8}{6}+4a+2b=16\end{array}\right}\\.\qquad\dfrac{6}{6}+2a=4\\\\.\qquad1+2a=4\qquad\text{subtract 1 from both sides}\\\\.\qquad2a=3\qquad\text{divide both sides by 2}\\\\.\qquad\boxed{a=\dfrac{3}{2}}[/tex]
Substitute it to (1):
[tex]\dfrac{1}{6}+\dfrac{3}{2}+b=6\qquad\text{multiply both sides by 6}\\\\(6)\left(\dfrac{1}{6}\right)+(6)\left(\dfrac{3}{2}\right)+6b=(6)(6)\\\\1+9+6b=36\\\\10+6b=36\qquad\text{subtract 10 from both sides}\\\\6b=26\qquad\text{divide both sides by 6}\\\\b=\dfrac{26}{6}\\\\\boxed{b=\dfrac{13}{3}}[/tex]
