QUESTION
Find the explicit formula for this sequence:
3,6,12,21,33,48,...
SOLUTION
notice that the difference between each term form an arithmetic sequence thus, it is a quadratic sequence.
[tex] \tt{equations} = \begin{cases} a + b + c = 3 & \text{} \\ 4a + 2b + c = 6 & \\9a + 3b + c = 12\text{}\end{cases}
[/tex]
Now,Subtract the equations to get the variables
[tex]\tt{Eq2 - Eq1} \\ (4a + 2b + c = 6) - (a + b + c = 3) \\ (4a + 2b + c = 6) + ( - a - b - c = - 3) \\ (4a - a + 2b - b + c - c = 6 - 3) \\ \green{ \large{(3a + b = 3)}}
[/tex]
[tex]\tt{Eq3 - Eq2}\\ (9a + 3b + c = 12) - (4a + 2b + c = 6) \\ (9a + 3b + c = 12) + ( - 4a - 2b - c = - 6) \\ (9a - 4a + 3b - 2b + c - c = 12 - 6) \\ \tt \large\green{(5a + b = 6)}[/tex]
Solve the system of equation
[tex]\tt5a + b = 6 \: \\ - 1(3a + b = - 3) \\ \\ \implies \: 5a + b = 6 \\ \: \: \: \: \: \: \: \: \: \: - 3a - b = 3 \\ \\ \implies \: 2a = 3 \\ \implies \: \large { \green{a = \boxed{\frac{3}{2} }}}[/tex]
Substitute the value of a to 3a+b=3
[tex]3a + b = 3 \\ b = 3 - 3 \green{( \frac{3}{2} )} \\ b = 3 - \frac{9}{2} \\ b = \green {\: \boxed{{ \frac{ - 3}{ \: \: 2} }}}[/tex]
Substitute again to get the value of c
[tex]a + b + c = 3 \\ \frac{3}{2} - \frac{3}{2} + c = 3 \\ c = \green{ \ \boxed{3}}[/tex]
Summing up the values to get the explicit formula
[tex]s(n) = an^{2} + bn + c \\ s(n) = \frac{3}{2} (n^{2}) - \frac{3}{2}n + 3 [/tex]
FINAL ANSWER
[tex] \green{ \large{s(n) = \frac{3}{2} n^{2} - \frac{3}{2}n + 3 }}[/tex]
