Find the solution of the recurrence relation: xn=3xn-1 + 1, where, x0=4
Let us find the solution of the recurrence relation: "x_n=3x_{n-1} + 1", where "x_0=4."
For this firstly, let us solve the characteristic equation of the homogeneous equation "x_n-3x_{n-1}=0" :
"k-3=0" or "k=3." It follows that the particular solution of the equation is "x_n=a=const", where "a=3a+1," and hence "a=-\\frac{1}{2}." The solution of the recurrence relation: "x_n=3x_{n-1} + 1"
is "x_n=C\\cdot3^n-\\frac{1}{2}." Taking into account that "x_0=4," we conclude that "4=x_0=C-\\frac{1}{2}", and thus "C=\\frac{9}{2}." Consequently,
"x_n=\\frac{9}{2}\\cdot3^n-\\frac{1}{2}=\\frac{3^{n+2}-1}{2}."
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