Question

What is the nth term rule of the quadratic sequence below? 4 , 6 , 10 , 16 , 24 , 34 , 46

Answer 1

Answer:

$\large\boxed{a_n=n^2-n+4}$

Step-by-step explanation:

$a_1=4,\ a_2=6,\ a_3=10,\ a_4=16,\ a_5=24,\ a_6=34,\ a_7=46\\\\a_n=an^2+bn+c\\\\for\ n=1\to 4=a(1^2)+b(1)+c\to a+c+b=4\\\\for\ n=2\to 6=a(2^2)+b(2)+c\to4a+2b+c=6\\\\for\ n=3\to3=a(3^2)+b(3)+c\to 9a+3b+c=10\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}a+b+c=4&(1)\\4a+2b+c=6&(2)\\9a+3b+c=10&(3)\end{array}\right\\\\a+b+c=4\qquad\text{subtract b and c from both sides}\\a=4-b-c\qquad\text{substitute to (2) and (3)}$

$(2)\\4(4-b-c)+2b+c=6\qquad\text{use the distributive property}\\(4)(4)+(4)(-b)+(4)(-c)+2b+c=6\\16-4b-4c+2b+c=6\qquad\text{subtract 16 from both sides}\\(-4b+2b)+(-4c+c)=-10\\-2b-3c=-10\\\\(3)\\9(4-b-c)+3b+c=10\qquad\text{use the distributive property}\\(9)(4)+(9)(-b)+(9)(-c)+3b+c=10\\36-9b-9c+3b+c=10\qquad\text{subtract 36 from both sides}\\(-9b+3b)+(-9c+c)=-26\\-6b-8c=-26$

$\text{Therefore we have the system of equations:}\\\left\{\begin{array}{ccc}-2b-3c=-10&\text{multiply both sides by (-3)}\\-6b-8c=-26\end{array}\right\\\underline{+\left\{\begin{array}{ccc}6b+9c=30\\-6b-8c=-26\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad\boxed{c=4}\\\\\text{Put the value of c to the first equation:}\\\\6b+9(4)=30\\6b+36=30\qquad\text{subtract 36 from both sides}\\6b=-6\qquad\text{divide both sides by 6}\\\boxed{b=-1}\\\\\text{Put the values of b and c to (1)}$

$a=4-(-1)-4\\a=4+1-4\\\boxed{a=1}\\\\\text{Therefore we have the formula of nth term:}\\\\a_n=1n^2-1n+4$