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1. The general term of the quadratic number pattern is \( a_n = 2n^2 - 3n \).
2. The 48th term of the quadratic number pattern is 4464.
3. The sum of the first differences of the quadratic number pattern is \( S_n = 2n^2 + n \).
Solución
Alright, I need to solve this problem involving a quadratic number pattern. The first four terms are given as -1, 2, 9, and 20. Let's break down the problem into parts and tackle each one step by step.
**2.1 Determine the general term of the quadratic number pattern**
First, I need to find the general term of the quadratic sequence. A quadratic sequence has the form:
\[ a_n = an^2 + bn + c \]
where \( a \), \( b \), and \( c \) are constants to be determined.
Given the first four terms:
- \( a_1 = -1 \)
- \( a_2 = 2 \)
- \( a_3 = 9 \)
- \( a_4 = 20 \)
Let's set up equations based on these terms.
For \( n = 1 \):
\[ a(1)^2 + b(1) + c = -1 \]
\[ a + b + c = -1 \] ...(1)
For \( n = 2 \):
\[ a(2)^2 + b(2) + c = 2 \]
\[ 4a + 2b + c = 2 \] ...(2)
For \( n = 3 \):
\[ a(3)^2 + b(3) + c = 9 \]
\[ 9a + 3b + c = 9 \] ...(3)
For \( n = 4 \):
\[ a(4)^2 + b(4) + c = 20 \]
\[ 16a + 4b + c = 20 \] ...(4)
Now, I have a system of four equations with three unknowns. To solve for \( a \), \( b \), and \( c \), I can use methods like substitution or elimination. Let's try elimination.
Subtract equation (1) from equation (2):
\[ (4a + 2b + c) - (a + b + c) = 2 - (-1) \]
\[ 3a + b = 3 \] ...(5)
Subtract equation (2) from equation (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 9 - 2 \]
\[ 5a + b = 7 \] ...(6)
Now, subtract equation (5) from equation (6):
\[ (5a + b) - (3a + b) = 7 - 3 \]
\[ 2a = 4 \]
\[ a = 2 \]
Now, substitute \( a = 2 \) back into equation (5):
\[ 3(2) + b = 3 \]
\[ 6 + b = 3 \]
\[ b = -3 \]
Finally, substitute \( a = 2 \) and \( b = -3 \) back into equation (1):
\[ 2 + (-3) + c = -1 \]
\[ -1 + c = -1 \]
\[ c = 0 \]
So, the general term of the quadratic sequence is:
\[ a_n = 2n^2 - 3n \]
**2.2 Calculate the value of the 48th term of the quadratic number pattern**
Using the general term \( a_n = 2n^2 - 3n \), let's find \( a_{48} \):
\[ a_{48} = 2(48)^2 - 3(48) \]
\[ a_{48} = 2(2304) - 144 \]
\[ a_{48} = 4608 - 144 \]
\[ a_{48} = 4464 \]
So, the 48th term is 4464.
**2.3 Show that the sum of the first differences of this quadratic number pattern can be given by \( S_n = 2n^2 + n \)**
First, let's recall that the first difference of a sequence is the difference between consecutive terms. For a quadratic sequence, the first differences themselves form an arithmetic sequence.
Given the general term \( a_n = 2n^2 - 3n \), let's find the first difference \( d_n = a_{n+1} - a_n \):
\[ d_n = [2(n+1)^2 - 3(n+1)] - [2n^2 - 3n] \]
\[ d_n = [2(n^2 + 2n + 1) - 3n - 3] - [2n^2 - 3n] \]
\[ d_n = 2n^2 + 4n + 2 - 3n - 3 - 2n^2 + 3n \]
\[ d_n = (2n^2 - 2n^2) + (4n - 3n + 3n) + (2 - 3) \]
\[ d_n = 4n - 1 \]
So, the first difference is \( d_n = 4n - 1 \).
Now, the sum of the first \( n \) differences \( S_n \) is:
\[ S_n = \sum_{k=1}^{n} d_k = \sum_{k=1}^{n} (4k - 1) \]
\[ S_n = 4\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 \]
\[ S_n = 4\left(\frac{n(n+1)}{2}\right) - n \]
\[ S_n = 2n(n+1) - n \]
\[ S_n = 2n^2 + 2n - n \]
\[ S_n = 2n^2 + n \]
Therefore, the sum of the first differences of the quadratic number pattern is \( S_n = 2n^2 + n \).
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