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To find the [tex]\( n \)[/tex]th term of the given quadratic sequence [tex]\( 3, 8, 15, 24, 35, \ldots \)[/tex], we'll follow a systematic approach. A quadratic sequence generally has the form:
[tex]\[ a n^2 + b n + c \][/tex]
We need to determine the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]. Here are the steps:
1. Identify the first differences:
- Calculate the differences between consecutive terms of the sequence.
[tex]\[ 8 - 3 = 5 \][/tex]
[tex]\[ 15 - 8 = 7 \][/tex]
[tex]\[ 24 - 15 = 9 \][/tex]
[tex]\[ 35 - 24 = 11 \][/tex]
Thus, the first differences are:
[tex]\[ 5, 7, 9, 11 \][/tex]
2. Identify the second differences:
- Calculate the differences between consecutive terms of the first differences.
[tex]\[ 7 - 5 = 2 \][/tex]
[tex]\[ 9 - 7 = 2 \][/tex]
[tex]\[ 11 - 9 = 2 \][/tex]
Since the second differences are constant (equal to 2), we confirm that the sequence is quadratic.
3. Form the system of equations:
- Use the first three terms of the sequence to set up equations and solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
Let’s use the terms for [tex]\( n = 1, 2, 3 \)[/tex]:
[tex]\[ T_1 = a(1)^2 + b(1) + c = 3 \quad \text{(Equation 1)} \][/tex]
[tex]\[ T_2 = a(2)^2 + b(2) + c = 8 \quad \text{(Equation 2)} \][/tex]
[tex]\[ T_3 = a(3)^2 + b(3) + c = 15 \quad \text{(Equation 3)} \][/tex]
These equations are:
[tex]\[ a + b + c = 3 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 4a + 2b + c = 8 \quad \text{(Equation 2)} \][/tex]
[tex]\[ 9a + 3b + c = 15 \quad \text{(Equation 3)} \][/tex]
4. Solve the system of equations:
Subtract Equation 1 from Equation 2:
[tex]\[ (4a + 2b + c) - (a + b + c) = 8 - 3 \][/tex]
[tex]\[ 3a + b = 5 \quad \text{(Equation 4)} \][/tex]
Subtract Equation 2 from Equation 3:
[tex]\[ (9a + 3b + c) - (4a + 2b + c) = 15 - 8 \][/tex]
[tex]\[ 5a + b = 7 \quad \text{(Equation 5)} \][/tex]
Subtract Equation 4 from Equation 5:
[tex]\[ (5a + b) - (3a + b) = 7 - 5 \][/tex]
[tex]\[ 2a = 2 \][/tex]
[tex]\[ a = 1 \][/tex]
Substitute [tex]\( a = 1 \)[/tex] back into Equation 4:
[tex]\[ 3(1) + b = 5 \][/tex]
[tex]\[ 3 + b = 5 \][/tex]
[tex]\[ b = 2 \][/tex]
Substitute [tex]\( a = 1 \)[/tex] and [tex]\( b = 2 \)[/tex] into Equation 1:
[tex]\[ 1 + 2 + c = 3 \][/tex]
[tex]\[ 3 + c = 3 \][/tex]
[tex]\[ c = 0 \][/tex]
5. Write the general formula for the nth term:
Using [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 0 \)[/tex], we get the nth term as:
[tex]\[ T_n = 1n^2 + 2n + 1 \implies T_n = n^2 + n + 1 \][/tex]
Thus, the nth term of the quadratic sequence [tex]\( 3, 8, 15, 24, 35, \ldots \)[/tex] is:
[tex]\[ n^2 + n + 1 \][/tex]
[tex]\[ a n^2 + b n + c \][/tex]
We need to determine the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]. Here are the steps:
1. Identify the first differences:
- Calculate the differences between consecutive terms of the sequence.
[tex]\[ 8 - 3 = 5 \][/tex]
[tex]\[ 15 - 8 = 7 \][/tex]
[tex]\[ 24 - 15 = 9 \][/tex]
[tex]\[ 35 - 24 = 11 \][/tex]
Thus, the first differences are:
[tex]\[ 5, 7, 9, 11 \][/tex]
2. Identify the second differences:
- Calculate the differences between consecutive terms of the first differences.
[tex]\[ 7 - 5 = 2 \][/tex]
[tex]\[ 9 - 7 = 2 \][/tex]
[tex]\[ 11 - 9 = 2 \][/tex]
Since the second differences are constant (equal to 2), we confirm that the sequence is quadratic.
3. Form the system of equations:
- Use the first three terms of the sequence to set up equations and solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
Let’s use the terms for [tex]\( n = 1, 2, 3 \)[/tex]:
[tex]\[ T_1 = a(1)^2 + b(1) + c = 3 \quad \text{(Equation 1)} \][/tex]
[tex]\[ T_2 = a(2)^2 + b(2) + c = 8 \quad \text{(Equation 2)} \][/tex]
[tex]\[ T_3 = a(3)^2 + b(3) + c = 15 \quad \text{(Equation 3)} \][/tex]
These equations are:
[tex]\[ a + b + c = 3 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 4a + 2b + c = 8 \quad \text{(Equation 2)} \][/tex]
[tex]\[ 9a + 3b + c = 15 \quad \text{(Equation 3)} \][/tex]
4. Solve the system of equations:
Subtract Equation 1 from Equation 2:
[tex]\[ (4a + 2b + c) - (a + b + c) = 8 - 3 \][/tex]
[tex]\[ 3a + b = 5 \quad \text{(Equation 4)} \][/tex]
Subtract Equation 2 from Equation 3:
[tex]\[ (9a + 3b + c) - (4a + 2b + c) = 15 - 8 \][/tex]
[tex]\[ 5a + b = 7 \quad \text{(Equation 5)} \][/tex]
Subtract Equation 4 from Equation 5:
[tex]\[ (5a + b) - (3a + b) = 7 - 5 \][/tex]
[tex]\[ 2a = 2 \][/tex]
[tex]\[ a = 1 \][/tex]
Substitute [tex]\( a = 1 \)[/tex] back into Equation 4:
[tex]\[ 3(1) + b = 5 \][/tex]
[tex]\[ 3 + b = 5 \][/tex]
[tex]\[ b = 2 \][/tex]
Substitute [tex]\( a = 1 \)[/tex] and [tex]\( b = 2 \)[/tex] into Equation 1:
[tex]\[ 1 + 2 + c = 3 \][/tex]
[tex]\[ 3 + c = 3 \][/tex]
[tex]\[ c = 0 \][/tex]
5. Write the general formula for the nth term:
Using [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 0 \)[/tex], we get the nth term as:
[tex]\[ T_n = 1n^2 + 2n + 1 \implies T_n = n^2 + n + 1 \][/tex]
Thus, the nth term of the quadratic sequence [tex]\( 3, 8, 15, 24, 35, \ldots \)[/tex] is:
[tex]\[ n^2 + n + 1 \][/tex]
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