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The firs four terms of a quadratic number pattem are: \( -1 ; 2 ; 9,20 \) 2.1 Determine the general term of the quadratic number pattert 2.2 Calculate the value of the \( 48^{\text {th }} \) teno of the quadiatic number pattern 2.3 Show that the sum of the first differences of this quadratic number pattem can be given by \( S_{n}=2 n^{2}+n \)

Ask by Ingram Patel. in South Africa
Mar 14,2025

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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 \( S_n = 2n^2 + n \).

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The Deep Dive

To determine the general term of the quadratic number pattern, consider that the quadratic expression can be represented as \( a_n = An^2 + Bn + C \). By plugging in the values for \( n=1, 2, 3, \) and \( 4 \) corresponding to the outputs \( -1, 2, 9, 20 \), you can create a system of equations to solve for coefficients \( A, B, \) and \( C \). Solving these equations gives \( a_n = n^2 + n - 1 \). Now for the \( 48^{\text{th}} \) term, simply substitute \( n=48 \) into the general term found: \( a_{48} = 48^2 + 48 - 1 \). This evaluates to \( 2304 + 48 - 1 = 2351 \). For the sum of the first differences, you will find that the first differences of the quadratic pattern are linear, represented as \( d_n = a_{n+1} - a_n = 2n + 1 \). Thus, the sum of the first \( n \) differences leads you to the formula \( S_n = 2n^2 + n \) through simple series summation techniques. This creates a delightful harmony between arithmetic and algebra!

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