Considre the sequence an n- 0 given by the following relation- a0-4-a1-22- and for all n- 2- an-6an-1-an-2- Prove that there exist sequences of positive integers xn n- 0- yn n- 0 such that an-y2n-7-xn-yn- for all n- 0-

Observe that a_n is an even positive integer for all n and that the sequence ( a_n )n≥ 0 is increasing. The last asssertion follows inductively if w ...

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Standard XII

Mathematics

Summation by Sigma Method

Considre the ...

Question

Considre the sequence $(a_{n}) n \geq 0$ given by the following relation: $a_{0} = 4, a_{1} = 22,$ and for all $n \geq 2,$ $a_{n} = 6 a_{n - 1} - a_{n - 2} .$ Prove that there exist sequences of positive integers $(x_{n}) n \geq 0, (y_{n}) n \geq 0$ such that $a_{n} = \frac{y_{n}^{2} + 7}{x_{n} - y_{n}},$ for all $n \geq 0.$

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Solution

Observe that $a_{n}$ is an even positive integer for all n and that the sequence $(a_{n}) n \geq 0$ is increasing. The last asssertion follows inductively if we write the recursive relation under the form $a_{n} - a_{n - 1} = 5 (a_{n - 1} - a_{n - 2}) + 4 a_{n - 2} .$ Define $x_{n} = \frac{a_{n} + a_{n - 1}}{2}, y_{n} = \frac{a_{n} - a_{n - 1}}{2}$ whith x0 = 3, y0 = 1. Observe that $x_{n} = 3 x_{n - 1} + 4 y_{n - 1}$ and $y_{n} = 2 x_{n - 1} + 3 y_{n - 1} .$ We have $a_{n} = x_{n} + y_{n},$ hence it is sufficient to prove that $x_{n} + y_{n} = \frac{y_{n}^{2} + 7}{x_{n} - y_{n}}$ or $x_{n}^{2} = 2 y_{n}^{2} + 7$ for all n. We use induction. The equality is true for n = 0. Suppose that $x_{n - 1}^{2} = 2 y_{n - 1}^{2} + 7.$ Then $x_{n}^{2} - 2 y_{n}^{2} - 7 = {(3 x_{n - 1} + 4 y_{n - 1})}_{n - 1}^{2} - 2 {(2 x_{n - 1 + 3 y_{n - 1}})}_{n - 1 + 3 y_{n - 1}}^{2} - 7$ $= x_{n - 1}^{2} - 2 y_{n - 1}^{2} - 7 = 0,$ which proves our claim. Try your hand at the following problems.

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Considre the sequence (an)n≥0 given by the following relation:a0=4,a1=22, and for all n≥2, an=6an−1−an−2. Prove that there exist sequences of positive integers (xn)n≥0,(yn)n≥0 such thatan=y2n+7xn−yn, for all n≥0.