Question

Se consideră numerele reale:
a1, a2, ..., an, ... astfel încât a1 = 2, a2= 5, an+2 = 5an+2 - 6an, n apartine N. n>=1. Să se arate că an= 2^n + 3^n, n apartine N.
URGENT!!!DAU COROANA!!!

Answer 1

Answer:

please send it in english then iwill answer

Answer 2

SOLUTION

TO SOLVE

The recurrence relation

$\sf{a_{n + 2} = 5a_{n + 1} - 6a_{n}}$

With initial conditions

$\sf{a_{1} = 2 \: \: \: and \: \: \: a_{2} = 5}$

EVALUATION

Here the given recurrence relation is

$\sf{a_{n + 2} = 5a_{n + 1} - 6a_{n}}$

$\implies \sf{a_{n + 2} - 5a_{n + 1} + 6a_{n} = 0}$

This is a second order homogeneous difference equation with constant coefficients

The corresponding auxiliary equation is

$\sf{ {x}^{2} - 5x + 6 = 0 }$

$\implies \sf{(x - 2)(x - 3) = 0}$

$\implies \sf{x = 2 \: , \: 3}$

Hence there e exists constants b and c such that

$\sf{a_{n} = b. {2}^{n} + c. {3}^{n} \: \: \: where \: \: n \geqslant 1 }$

Now it is given that

$\sf{a_{1} = 2 \: \: \: and \: \: \: a_{2} = 5}$

Here

$\sf{a_{1} = 2 \: \: \: gives}$

$\sf{2b + 3c = 2} \: \: \: \: ......(1)$

$\sf{a_{2} = 5 \: \: \: gives}$

$\sf{4b + 9c = 5} \: \: \: \: ......(2)$

Equation (2) - 2 × Equation (1) gives

$\sf{3c = 1}$

$\displaystyle \sf{ \implies \: c = \frac{1}{3} }$

From Equation (1) we get

$\sf{2b + 1 = 2}$

$\displaystyle \sf{ \implies \: 2b = 1}$

$\displaystyle \sf{ \implies \: b = \frac{1}{2} }$

Hence the required recurrence relation is

$\displaystyle \: \sf{a_{n} = \bigg( \frac{1}{2} \times {2}^{n} + \frac{1}{3} \times {3}^{n} \bigg) \: \: \: \: \: where \: \: n \geqslant 1 }$

$\displaystyle \: \sf{ \therefore \: \: \: a_{n} = {2}^{n - 1} + {3}^{n - 1} \: \: \: \: \: where \: \: n \geqslant 1 }$

FINAL ANSWER

The required recurrence relation is

$\displaystyle \: \sf{ a_{n} = {2}^{n - 1} + {3}^{n - 1} \: \: \: \: \: where \: \: n \geqslant 1 }$

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