Example Question - arithmetic progression
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Recursive Sequence Calculation
Pour la première suite : <p>u_0 = 2</p> <p>u_{n+1} = 3u_n - 4n</p> <p>u_1 = 3u_0 - 4 \cdot 0 = 3 \cdot 2 - 0 = 6</p> <p>u_2 = 3u_1 - 4 \cdot 1 = 3 \cdot 6 - 4 = 18 - 4 = 14</p> <p>u_3 = 3u_2 - 4 \cdot 2 = 3 \cdot 14 - 8 = 42 - 8 = 34</p> Pour la deuxième suite : <p>u_0 = 0</p> <p>u_{n+1} = u_n^2 + \frac{1}{2n + 1}</p> <p>u_1 = u_0^2 + \frac{1}{2 \cdot 0 + 1} = 0^2 + \frac{1}{1} = 0 + 1 = 1</p> <p>u_2 = u_1^2 + \frac{1}{2 \cdot 1 + 1} = 1^2 + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3}</p> <p>u_3 = u_2^2 + \frac{1}{2 \cdot 2 + 1} = \left(\frac{4}{3}\right)^2 + \frac{1}{5} = \frac{16}{9} + \frac{1}{5} = \frac{80}{45} + \frac{9}{45} = \frac{89}{45}</p>
Recurrence Relations in Sequences
Pour la 1ère suite : <p> \( u_0 = 2 \) </p> <p> \( u_1 = 3u_0 - 4 \times 0 = 3 \times 2 - 0 = 6 \) </p> <p> \( u_2 = 3u_1 - 4 \times 1 = 3 \times 6 - 4 = 18 - 4 = 14 \) </p> <p> \( u_3 = 3u_2 - 4 \times 2 = 3 \times 14 - 8 = 42 - 8 = 34 \) </p> Pour la 2ème suite : <p> \( u_0 = 0 \) </p> <p> \( u_1 = u_0^2 + \frac{1}{2 \times 0 + 1} = 0^2 + \frac{1}{1} = 0 + 1 = 1 \) </p> <p> \( u_2 = u_1^2 + \frac{1}{2 \times 1 + 1} = 1^2 + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3} \) </p> <p> \( u_3 = u_2^2 + \frac{1}{2 \times 2 + 1} = \left(\frac{4}{3}\right)^2 + \frac{1}{5} = \frac{16}{9} + \frac{1}{5} = \frac{80}{45} + \frac{9}{45} = \frac{89}{45} \) </p>
Sequence Calculation Problem
<p>Le premier terme de la suite est donné par \( u_0 = 2 \).</p> <p>Pour calculer \( u_1 \), on utilise la formule récurrente : \( u_1 = 3u_0 - 4n \) avec \( n=0 \), donc : \( u_1 = 3(2) - 4(0) = 6 \).</p> <p>Pour calculer \( u_2 \), on utilise la même formule récurrente : \( u_2 = 3u_1 - 4(1) = 3(6) - 4 = 18 - 4 = 14 \).</p> <p>Pour calculer \( u_3 \), on continue avec la formule récurrente : \( u_3 = 3u_2 - 4(2) = 3(14) - 8 = 42 - 8 = 34 \).</p> <p>Donc, les trois termes suivants le premier terme \( u_0 = 2 \) sont \( u_1 = 6 \), \( u_2 = 14 \), et \( u_3 = 34 \).</p>
Solving a Sequence Problem
<p>На данном изображении приведена задача на арифметическую последовательность, в которой необходимо найти сумму первых n членов.</p> <p>Формула суммы первых n членов арифметической прогрессии: \( S_n = \frac{n}{2} \cdot (a_1 + a_n) \), где \( S_n \) - сумма первых n членов, \( a_1 \) - первый член, и \( a_n \) - n-ый член прогрессии.</p> <p>Сначала найдем разность прогрессии (d):</p> <p>\( a_2 = a_1 + d \)</p> <p>\( 6,1 = 5,7 + d \)</p> <p>\( d = 6,1 - 5,7 \)</p> <p>\( d = 0,4 \)</p> <p>Теперь найдем 14-ый член прогрессии \( a_{14} \):</p> <p>\( a_{14} = a_1 + (14 - 1) \cdot d \)</p> <p>\( a_{14} = 5,7 + 13 \cdot 0,4 \)</p> <p>\( a_{14} = 5,7 + 5,2 \)</p> <p>\( a_{14} = 10,9 \)</p> <p>Теперь мы можем вычислить сумму первых 14 членов ( \( S_{14} \) ):</p> <p>\( S_{14} = \frac{14}{2} \cdot (a_1 + a_{14}) \)</p> <p>\( S_{14} = 7 \cdot (5,7 + 10,9) \)</p> <p>\( S_{14} = 7 \cdot 16,6 \)</p> <p>\( S_{14} = 116,2 \)</p> <p>Таким образом, сумма первых 14 членов данной арифметической прогрессии равна 116,2.</p>
Finding the Eighth Term in an Arithmetic Progression
<p>The first term of an arithmetic progression (AP) is given as \( a_1 = 3 \).</p> <p>The sum of the first and sixth terms is 20. We can express the sixth term as \( a_6 = a_1 + 5d \), where \( d \) is the common difference of the AP.</p> <p>Therefore, \( a_1 + a_6 = 20 \).</p> <p>Substitute \( a_1 \) and \( a_6 \) with given values:</p> <p>\( 3 + (3 + 5d) = 20 \)</p> <p>Solve for \( d \):</p> <p>\( 6 + 5d = 20 \)</p> <p>\( 5d = 14 \)</p> <p>\( d = \frac{14}{5} \)</p> <p>Now, find the eighth term \( a_8 \) using \( a_8 = a_1 + 7d \):</p> <p>\( a_8 = 3 + 7\left(\frac{14}{5}\right) \)</p> <p>\( a_8 = 3 + \frac{98}{5} \)</p> <p>\( a_8 = 3 + 19.6 \)</p> <p>\( a_8 = 22.6 \)</p>
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