Solve each of the following recurrence equations with the given
initial values.
(a) bn = bn-1 + 12bn-2. Initial values: b0 = -2, b1
= 20.
(b) bn = 3bn-1 + 4bn-2. Initial values: b0 = 4, b1
= 1.
(c) bn = 4bn-2. Initial values: b0 = 2, b1 =
16.
(d) bn = 4bn-1 - 4bn-2. Initial values: b0 = 3, b1
= 10.
(e) bn = 3bn-1 + 4bn-2 - 12bn-3. Initial values: b0
= 4, b1 = -5, b2 = 11.
(f) bn = 5bn-1 - 8bn-2 + 4bn-3. Initial values: b0
= 6, b1 = 7, b2 = 17.

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(a)b_(n)=b_(n-1)+12b_(n-2);b_(0)=-2,b_(1)=20The characteristic polynomial is x^(2)-x-12 having roots -3 and 4 .The general solution is b_(n)=C_(1)(-3)^(n)+C_(2)(4)^(n)For the values of n as 0 and 1 , the two different equations are as follows:{:[C_(1)+C_(2)=-2],[-3C_(1)+4C_(2)=20]:}[Using: Initial values: b0 =-2,b1=20 ]Solving the above equations, the values of C_(1) and C_(2) are as follows:{:[C_(1)=-4],[C_(2)=2]:}So, the solution is b_(n)=(-4)(-3)^(n)+2(4)^(n)b)b_(n)=3b_(n-1)+4b_(n-2);b_(0)=4,b_(1)=1The characteristic polynomial is x^(2)-3x-4 having roots -1 and 4 .The general solution is b_(n)=C_(1)(-1)^(n)+C_(2)(4)^(n)For the initial values of n as 0 and 1 , the two different equations are as follows:{:[C_(1)+C_(2)=4],[-C_(1)+4C_(2)=1]:}On solving the above equations, the values of C_(1) and C_(2) are as follows:{:[C_(1)=3],[C_(2)=1]:}The ... See the full answer