Pregunta

Let \( T_{n}=4 n^{2}+n \) be the general term of a quadratic sequence with the fourth term being 68. 21 and 29 are the second and third terms, respectively, of the sequence of the first differences of the quadratic sequence.

Ask by Ingram Wells. in South Africa
Jan 24,2025

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The quadratic sequence \( T_n = 4n^2 + n \) has a fourth term of 68. The second and third terms of the first differences are 21 and 29, respectively.

Solución

To solve the problem, we need to analyze the quadratic sequence given by the general term \( T_n = 4n^2 + n \). ### Step 1: Find the fourth term \( T_4 \) We can calculate the fourth term using the formula for \( T_n \): \[ T_4 = 4(4^2) + 4 = 4(16) + 4 = 64 + 4 = 68 \] This confirms that the fourth term is indeed 68. ### Step 2: Find the first differences of the sequence The first differences of a sequence are found by calculating \( T_{n+1} - T_n \). Let's calculate the first differences for the quadratic sequence: \[ T_{n+1} = 4(n+1)^2 + (n+1) = 4(n^2 + 2n + 1) + n + 1 = 4n^2 + 8n + 4 + n + 1 = 4n^2 + 9n + 5 \] Now, we can find the first difference: \[ T_{n+1} - T_n = (4n^2 + 9n + 5) - (4n^2 + n) = 8n + 5 \] ### Step 3: Calculate the first few first differences Now we can calculate the first differences for the first few terms: - For \( n = 1 \): \[ T_2 - T_1 = 8(1) + 5 = 8 + 5 = 13 \] - For \( n = 2 \): \[ T_3 - T_2 = 8(2) + 5 = 16 + 5 = 21 \] - For \( n = 3 \): \[ T_4 - T_3 = 8(3) + 5 = 24 + 5 = 29 \] ### Step 4: Verify the first differences From our calculations, we have: - The first difference for \( n = 1 \) is 13. - The first difference for \( n = 2 \) is 21. - The first difference for \( n = 3 \) is 29. The second term of the first differences is 21, and the third term is 29, which matches the conditions given in the problem. ### Conclusion The quadratic sequence defined by \( T_n = 4n^2 + n \) has the fourth term as 68, and the second and third terms of the first differences are indeed 21 and 29, respectively. Thus, all conditions of the problem are satisfied.

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The general term of a quadratic sequence is given by \( T_n = 4n^2 + n \). To find the fourth term, we can substitute \( n=4 \): \[ T_4 = 4(4^2) + 4 = 4(16) + 4 = 64 + 4 = 68 \] This confirms the fourth term is indeed 68. Now, the first differences of the quadratic sequence are calculated by finding \( T_{n+1} - T_n \). Expressing this calculation gives us a linear sequence since the first differences of quadratic sequences are linear. The first term is given by \( T_1 = 4(1^2) + 1 = 5 \). The second term is \( T_2 = 4(2^2) + 2 = 18 \). The third term is \( T_3 = 4(3^2) + 3 = 39 \). The first differences are \( T_2 - T_1 = 18 - 5 = 13 \) (first difference), \( T_3 - T_2 = 39 - 18 = 21 \) (second difference), and \( T_4 - T_3 = 68 - 39 = 29 \) (third difference). This leads to the first differences being \( 13, 21, 29, ...\). The pattern in the first differences helps to reveal that each difference is increasing by 8, consistent with its quadratic nature. In terms of historical context, quadratic sequences have fascinated mathematicians since ancient times. Notably, they relate to polynomial growth, which is foundational not only in theoretical mathematics but also in areas like physics and economics, where relationships are often modeled using quadratic equations. For practical applications, understanding quadratic sequences can help decipher various real-world problems. For instance, in projectile motion, the height of an object at a specific time can be modeled as a quadratic equation. Knowing how to derive and work with these sequences allows for predictions about the object's behavior at various time intervals, assisting engineers and physicists in designs and calculations.

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