Answer :
To find the nth term of the given quadratic sequence 3, 8, 15, 24, 35, ..., follow these steps:
### Step 1: Determine the Sequence Type
The first differences between consecutive terms help identify the sequence type:
[tex]\[ \begin{aligned} 8 - 3 &= 5 \\ 15 - 8 &= 7 \\ 24 - 15 &= 9 \\ 35 - 24 &= 11 \\ \end{aligned} \][/tex]
The first differences are: 5, 7, 9, 11.
Calculate the second differences:
[tex]\[ \begin{aligned} 7 - 5 &= 2 \\ 9 - 7 &= 2 \\ 11 - 9 &= 2 \\ \end{aligned} \][/tex]
The second differences are constant: 2. Thus, we have a quadratic sequence.
### Step 2: General Form of a Quadratic Sequence
The nth term of a quadratic sequence can be represented as:
[tex]\[ a_n = An^2 + Bn + C \][/tex]
### Step 3: Setup and Solve the Equations
We need to find the coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] given the first few terms of the sequence. Set up equations using the given terms:
For [tex]\(n = 1\)[/tex]:
[tex]\[ A(1)^2 + B(1) + C = 3 \][/tex]
[tex]\[ A + B + C = 3 \][/tex]
For [tex]\(n = 2\)[/tex]:
[tex]\[ A(2)^2 + B(2) + C = 8 \][/tex]
[tex]\[ 4A + 2B + C = 8 \][/tex]
For [tex]\(n = 3\)[/tex]:
[tex]\[ A(3)^2 + B(3) + C = 15 \][/tex]
[tex]\[ 9A + 3B + C = 15 \][/tex]
Now, we have the system of equations:
[tex]\[ \begin{aligned} 1. & \quad A + B + C = 3 \\ 2. & \quad 4A + 2B + C = 8 \\ 3. & \quad 9A + 3B + C = 15 \\ \end{aligned} \][/tex]
### Step 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]
Finally, substitute [tex]\(A = 1\)[/tex] and [tex]\(B = 2\)[/tex] back into equation 1:
[tex]\[ 1 + 2 + C = 3 \][/tex]
[tex]\[ 3 + C = 3 \][/tex]
[tex]\[ C = 0 \][/tex]
### Step 5: Formulate the nth Term
With [tex]\(A = 1\)[/tex], [tex]\(B = 2\)[/tex], and [tex]\(C = 0\)[/tex], the nth term of the sequence is:
[tex]\[ a_n = n^2 + 2n \][/tex]
Thus, the nth term of the quadratic sequence [tex]\(3, 8, 15, 24, 35, \ldots\)[/tex] is:
[tex]\[ \boxed{a_n = n^2 + 2n} \][/tex]
### Step 1: Determine the Sequence Type
The first differences between consecutive terms help identify the sequence type:
[tex]\[ \begin{aligned} 8 - 3 &= 5 \\ 15 - 8 &= 7 \\ 24 - 15 &= 9 \\ 35 - 24 &= 11 \\ \end{aligned} \][/tex]
The first differences are: 5, 7, 9, 11.
Calculate the second differences:
[tex]\[ \begin{aligned} 7 - 5 &= 2 \\ 9 - 7 &= 2 \\ 11 - 9 &= 2 \\ \end{aligned} \][/tex]
The second differences are constant: 2. Thus, we have a quadratic sequence.
### Step 2: General Form of a Quadratic Sequence
The nth term of a quadratic sequence can be represented as:
[tex]\[ a_n = An^2 + Bn + C \][/tex]
### Step 3: Setup and Solve the Equations
We need to find the coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] given the first few terms of the sequence. Set up equations using the given terms:
For [tex]\(n = 1\)[/tex]:
[tex]\[ A(1)^2 + B(1) + C = 3 \][/tex]
[tex]\[ A + B + C = 3 \][/tex]
For [tex]\(n = 2\)[/tex]:
[tex]\[ A(2)^2 + B(2) + C = 8 \][/tex]
[tex]\[ 4A + 2B + C = 8 \][/tex]
For [tex]\(n = 3\)[/tex]:
[tex]\[ A(3)^2 + B(3) + C = 15 \][/tex]
[tex]\[ 9A + 3B + C = 15 \][/tex]
Now, we have the system of equations:
[tex]\[ \begin{aligned} 1. & \quad A + B + C = 3 \\ 2. & \quad 4A + 2B + C = 8 \\ 3. & \quad 9A + 3B + C = 15 \\ \end{aligned} \][/tex]
### Step 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]
Finally, substitute [tex]\(A = 1\)[/tex] and [tex]\(B = 2\)[/tex] back into equation 1:
[tex]\[ 1 + 2 + C = 3 \][/tex]
[tex]\[ 3 + C = 3 \][/tex]
[tex]\[ C = 0 \][/tex]
### Step 5: Formulate the nth Term
With [tex]\(A = 1\)[/tex], [tex]\(B = 2\)[/tex], and [tex]\(C = 0\)[/tex], the nth term of the sequence is:
[tex]\[ a_n = n^2 + 2n \][/tex]
Thus, the nth term of the quadratic sequence [tex]\(3, 8, 15, 24, 35, \ldots\)[/tex] is:
[tex]\[ \boxed{a_n = n^2 + 2n} \][/tex]
