Answer :
To find the \( n \)th term of the quadratic sequence \( 3, 8, 15, 24, 35, \ldots \), we assume the general form of a quadratic sequence, which is given by:
[tex]\[ T_n = an^2 + bn + c \][/tex]
Using the terms of the given sequence, we can set up a system of equations based on the first few terms.
1. For \( T_1 = 3 \), we have:
[tex]\[ a(1)^2 + b(1) + c = 3 \][/tex]
[tex]\[ a + b + c = 3 \quad \text{(Equation 1)} \][/tex]
2. For \( T_2 = 8 \), we have:
[tex]\[ a(2)^2 + b(2) + c = 8 \][/tex]
[tex]\[ 4a + 2b + c = 8 \quad \text{(Equation 2)} \][/tex]
3. For \( T_3 = 15 \), we have:
[tex]\[ a(3)^2 + b(3) + c = 15 \][/tex]
[tex]\[ 9a + 3b + c = 15 \quad \text{(Equation 3)} \][/tex]
We now have a system of three equations:
1. \( a + b + c = 3 \)
2. \( 4a + 2b + c = 8 \)
3. \( 9a + 3b + c = 15 \)
By solving this system of equations, we find the values of \( a \), \( b \), and \( c \). The calculations yield the following results:
[tex]\[ a \approx 1 \][/tex]
[tex]\[ b \approx 2 \][/tex]
[tex]\[ c \approx 0 \][/tex]
Therefore, the formula for the \( n \)th term of the sequence is:
[tex]\[ T_n = n^2 + 2n + 0 \][/tex]
or simply:
[tex]\[ T_n = n^2 + 2n \][/tex]
Let's verify by calculating the first few terms using this formula:
For \( n = 1 \):
[tex]\[ T_1 = 1^2 + 2 \cdot 1 = 1 + 2 = 3 \][/tex]
For \( n = 2 \):
[tex]\[ T_2 = 2^2 + 2 \cdot 2 = 4 + 4 = 8 \][/tex]
For \( n = 3 \):
[tex]\[ T_3 = 3^2 + 2 \cdot 3 = 9 + 6 = 15 \][/tex]
For \( n = 4 \):
[tex]\[ T_4 = 4^2 + 2 \cdot 4 = 16 + 8 = 24 \][/tex]
For \( n = 5 \):
[tex]\[ T_5 = 5^2 + 2 \cdot 5 = 25 + 10 = 35 \][/tex]
Hence, the \( n \)th term of the given quadratic sequence is:
[tex]\[ T_n = n^2 + 2n \][/tex]
[tex]\[ T_n = an^2 + bn + c \][/tex]
Using the terms of the given sequence, we can set up a system of equations based on the first few terms.
1. For \( T_1 = 3 \), we have:
[tex]\[ a(1)^2 + b(1) + c = 3 \][/tex]
[tex]\[ a + b + c = 3 \quad \text{(Equation 1)} \][/tex]
2. For \( T_2 = 8 \), we have:
[tex]\[ a(2)^2 + b(2) + c = 8 \][/tex]
[tex]\[ 4a + 2b + c = 8 \quad \text{(Equation 2)} \][/tex]
3. For \( T_3 = 15 \), we have:
[tex]\[ a(3)^2 + b(3) + c = 15 \][/tex]
[tex]\[ 9a + 3b + c = 15 \quad \text{(Equation 3)} \][/tex]
We now have a system of three equations:
1. \( a + b + c = 3 \)
2. \( 4a + 2b + c = 8 \)
3. \( 9a + 3b + c = 15 \)
By solving this system of equations, we find the values of \( a \), \( b \), and \( c \). The calculations yield the following results:
[tex]\[ a \approx 1 \][/tex]
[tex]\[ b \approx 2 \][/tex]
[tex]\[ c \approx 0 \][/tex]
Therefore, the formula for the \( n \)th term of the sequence is:
[tex]\[ T_n = n^2 + 2n + 0 \][/tex]
or simply:
[tex]\[ T_n = n^2 + 2n \][/tex]
Let's verify by calculating the first few terms using this formula:
For \( n = 1 \):
[tex]\[ T_1 = 1^2 + 2 \cdot 1 = 1 + 2 = 3 \][/tex]
For \( n = 2 \):
[tex]\[ T_2 = 2^2 + 2 \cdot 2 = 4 + 4 = 8 \][/tex]
For \( n = 3 \):
[tex]\[ T_3 = 3^2 + 2 \cdot 3 = 9 + 6 = 15 \][/tex]
For \( n = 4 \):
[tex]\[ T_4 = 4^2 + 2 \cdot 4 = 16 + 8 = 24 \][/tex]
For \( n = 5 \):
[tex]\[ T_5 = 5^2 + 2 \cdot 5 = 25 + 10 = 35 \][/tex]
Hence, the \( n \)th term of the given quadratic sequence is:
[tex]\[ T_n = n^2 + 2n \][/tex]
