To find the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the quadratic sequence [tex]\( x_n = an^2 + b \)[/tex], let's use the given terms of the sequence: 2, 6, 12, and 20. Each term corresponds to particular values of [tex]\( n \)[/tex]. Specifically:
- When [tex]\( n = 1 \)[/tex], [tex]\( x_1 = 2 \)[/tex]
- When [tex]\( n = 2 \)[/tex], [tex]\( x_2 = 6 \)[/tex]
- When [tex]\( n = 3 \)[/tex], [tex]\( x_3 = 12 \)[/tex]
- When [tex]\( n = 4 \)[/tex], [tex]\( x_4 = 20 \)[/tex]
Given the formula [tex]\( x_n = an^2 + b \)[/tex], we can set up the following equations by substituting the corresponding [tex]\( n \)[/tex] and [tex]\( x_n \)[/tex] values:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ a(1)^2 + b = 2 \][/tex]
[tex]\[ a + b = 2 \quad (1) \][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ a(2)^2 + b = 6 \][/tex]
[tex]\[ 4a + b = 6 \quad (2) \][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ a(3)^2 + b = 12 \][/tex]
[tex]\[ 9a + b = 12 \quad (3) \][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ a(4)^2 + b = 20 \][/tex]
[tex]\[ 16a + b = 20 \quad (4) \][/tex]
Now, we use equations (1) and (2) to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
From equation (1):
[tex]\[ b = 2 - a \quad (5) \][/tex]
Substitute equation (5) into equation (2):
[tex]\[ 4a + (2 - a) = 6 \][/tex]
[tex]\[ 4a + 2 - a = 6 \][/tex]
[tex]\[ 3a + 2 = 6 \][/tex]
[tex]\[ 3a = 4 \][/tex]
[tex]\[ a = \frac{4}{3} \][/tex]
Now, substitute [tex]\( a = \frac{4}{3} \)[/tex] back into equation (5):
[tex]\[ b = 2 - \frac{4}{3} \][/tex]
[tex]\[ b = \frac{6}{3} - \frac{4}{3} \][/tex]
[tex]\[ b = \frac{2}{3} \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are:
[tex]\[ a = \frac{4}{3} \][/tex]
[tex]\[ b = \frac{2}{3} \][/tex]
