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3.1 Consider the quadratic number pattern: \( 3 ; 7 ; 12 ; \ldots \) Show that the general term of this number patten is given by \( T_{n}=\frac{1}{2} n^{2}+\frac{5}{2} \) ?

Ask by Paul Barnett. in South Africa
Feb 21,2025

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The general term of the sequence is \( T_{n} = \frac{1}{2}n^{2} + \frac{5}{2} \).

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We are given the number pattern 3, 7, 12, … and are to show that its general term is of the form   Tₙ = (1/2)n² + (5/2)n. Step 1. Assume a quadratic form for Tₙ:   Tₙ = a·n² + b·n + c. Step 2. Use the first three terms to set up equations. For n = 1:   a(1)² + b(1) + c = a + b + c = 3      (Equation 1) For n = 2:   a(2)² + b(2) + c = 4a + 2b + c = 7     (Equation 2) For n = 3:   a(3)² + b(3) + c = 9a + 3b + c = 12    (Equation 3) Step 3. Subtract Equation 1 from Equation 2 to eliminate c:   (4a + 2b + c) − (a + b + c) = 7 − 3   3a + b = 4                  (Equation 4) Subtract Equation 2 from Equation 3:   (9a + 3b + c) − (4a + 2b + c) = 12 − 7   5a + b = 5                  (Equation 5) Step 4. Subtract Equation 4 from Equation 5 to solve for a:   (5a + b) − (3a + b) = 5 − 4   2a = 1   a = 1/2 Step 5. Substitute a = 1/2 back into Equation 4 to solve for b:   3(1/2) + b = 4   3/2 + b = 4   b = 4 − 3/2 = (8/2 − 3/2) = 5/2 Step 6. Substitute a = 1/2 and b = 5/2 back into Equation 1 to solve for c:   (1/2) + (5/2) + c = 3   (6/2) + c = 3   3 + c = 3   c = 0 Thus, the general term becomes:   Tₙ = (1/2)n² + (5/2)n. Step 7. Verify the formula: For n = 1:   T₁ = (1/2)(1)² + (5/2)(1) = 1/2 + 5/2 = 6/2 = 3. For n = 2:   T₂ = (1/2)(4) + (5/2)(2) = 2 + 5 = 7. For n = 3:   T₃ = (1/2)(9) + (5/2)(3) = 4.5 + 7.5 = 12. Since the values match the given pattern, we have shown that   Tₙ = (1/2)n² + (5/2)n correctly represents the sequence.

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To derive the general term \( T_n = \frac{1}{2} n^2 + \frac{5}{2} \) for the given pattern, we first observe the differences between consecutive terms: \( 7 - 3 = 4 \) and \( 12 - 7 = 5 \). The first differences are \( 4 \) and \( 5 \), suggesting the second differences are constant (which equals \( 1 \)). This is indicative of a quadratic sequence. Assuming a quadratic formula \( T_n = an^2 + bn + c \), we can set up a system using the initial conditions: when \( n=1, T_1=3 \); when \( n=2, T_2=7 \); and when \( n=3, T_3=12 \). Solving this system provides the coefficients \( a = \frac{1}{2}, b = 0, c = \frac{5}{2} \). Therefore, we can confirm that the general term is indeed \( T_n = \frac{1}{2} n^2 + \frac{5}{2} \). To visualize this quadratic progression, imagine plotting the points on a graph! The term values \( (1,3), (2,7), (3,12) \) form a parabolic curve. That classic upward curvature is a telltale sign of quadratics in the mathematical world, revealing how they grow faster than linear sequences. So, next time you see a pattern like this, think of it as a fun curve waiting to be explored! Also, if you're diving deeper into quadratic functions, consider getting into polynomial growth rates. They're not just for textbooks; these forms are sprinkled throughout nature, economics, and engineering! From predicting rocket trajectories to calculating maximum profit in business models, understanding quadratics can turn you into a mathematical wizard in real-world scenarios.

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