Answer:
[tex] \displaystyle a_{n} = (2)^{2n -1} - (3) ^{n-1 }[/tex]
Step-by-step explanation:
we want to figure out the general term of the following recurrence relation
[tex] \displaystyle \rm a_{n + 2} - 7a_{n + 1} + 12a_n = 0 \: \: where : \: \:a_1 = 1 \: ,a_2 = 5, [/tex]
we are given a linear homogeneous recurrence relation which degree is 2. In order to find the general term ,we need to make it a characteristic equation i.e
the steps for solving a linear homogeneous recurrence relation are as follows:
Step-1:Create the characteristic equation
[tex] {x}^{2} - 7x+ 12= 0[/tex]
Step-2:Solve the polynomial by factoring
factor the quadratic:
[tex]( {x}^{} - 4)(x - 3) = 0[/tex]
solve for x:
[tex]x = \rm 4 \:and \: 3[/tex]
Step-3:Determine the form for each solution
since we've two distinct roots,we'd utilize the following formula:
[tex] \displaystyle a_{n} = c_{1} {x} _{1} ^{n } + c_{2} {x} _{2} ^{n } [/tex]
so substitute the roots we got:
[tex] \displaystyle a_{n} = c_{1} (4)^{n } + c_{2} (3) ^{n } [/tex]
Step-4:Use initial conditions to find coefficients using systems of equations
create the system of equation:
[tex] \begin{cases}\displaystyle 4c_{1} +3 c_{2} = 1 \\ 16c_{1} + 9c_{2} = 5\end{cases}[/tex]
solve the system of equation which yields:
[tex] \displaystyle c_{1} = \frac{1}{2} \\ c_{2} = - \frac{1}{3} [/tex]
finally substitute:
[tex] \displaystyle a_{n} = \frac{1}{2} (4)^{n } - \frac{1}{3} (3) ^{n }[/tex]
[tex] \displaystyle \boxed{ a_{n} = (2)^{2n-1 } - (3) ^{n -1}}[/tex]
and we're done!
Consider the generating function of the sequence [tex]a_n[/tex], defined by
[tex]A(z) = \displaystyle \sum_{n=0}^\infty a_nz^n = a_0 + a_1z + a_2z^2 + \cdots[/tex]
(this is also known as the Z-transform of [tex]a_n[/tex])
While the given sequence isn't exactly defined for n = 0, we can use the recurrence to extend it to cover this case. For n = 0, we get
[tex]a_2 - 7a_1 + 12a_0 = 0 \implies a_0 = -\dfrac16[/tex]
Since for all n ∈ {0, 1, 2, …} we have that
[tex]a_{n+2}-7a_{n+1}+12a_n=0[/tex]
Multiplying both sides by [tex]z^n[/tex] and summing over all n gives the relation
[tex]\displaystyle \sum_{n=0}^\infty a_{n+2}z^n - 7 \sum_{n=0}^\infty a_{n+1}z^n + 12 \sum_{n=0}^\infty a_nz^n = 0[/tex]
Manipulate the first two series to get them in terms of A(z) :
[tex]\displaystyle \sum_{n=0}^\infty a_{n+2}z^n = \frac1{z^2}\sum_{n=0}^\infty a_{n+2}z^{n+2} \\\\ \sum_{n=0}^\infty a_{n+2}z^n = \frac1{z^2} \sum_{n=2}^\infty a_nz^n \\\\ \sum_{n=0}^\infty a_{n+2}z^n = \frac1{z^2}\left(\sum_{n=0}^\infty a_nz^n - a_0 - a_1z\right) \\\\ \sum_{n=0}^\infty a_{n+2}z^n = \frac{A(z)-\frac16-z}{z^2}[/tex]
and
[tex]\displaystyle \sum_{n=0}^\infty a_{n+1}z^n = \frac1z \sum_{n=0}^\infty a_{n+1}z^{n+1} \\\\ \sum_{n=0}^\infty a_{n+1}z^n = \frac1z\sum_{n=1}^\infty a_nz^n \\\\ \sum_{n=0}^\infty a_{n+1}z^n = \frac1z\left(\sum_{n=0}^\infty a_nz^n - a_0\right) \\\\ \sum_{n=0}^\infty a_{n+1}z^n = \frac{A(z)-\frac16}{z}[/tex]
The third series is simply A(z).
Solve this new recurrence for A(z) :
[tex]\displaystyle \frac{A(z)-\frac16-z}{z^2} - \frac{7A(z)-\frac76}{z} + 12 A(z) = 0 \\\\ \frac{A(z)}{z^2}-\frac{7A(z)}{z} + 12A(z) = \frac1{6z^2}+\frac1z-\frac7{6z} \\\\ \frac{A(z)-7zA(z)+12z^2A(z)}{z^2} = \frac{1-z}{6z^2} \\\\ \frac{1-7z+12z^2}{z^2}A(z) = \frac{1-z}{6z^2} \\\\ A(z) = \frac{1-z}{6(1-7z+12z^2)}[/tex]
The next step is to find the power series expansion for A(z) to recover [tex]a_n[/tex]. Factorize the denominator, then decompose A(z) into partial fractions:
[tex]1-7z+12z^2 = (1-3z)(1-4z)[/tex]
[tex]\displaystyle \frac{1-z}{(1-3z)(1-4z)} = \frac a{1-3z} + \frac b{1-4z} \\\\ 1-z = a(1-4z) + b(1-3z) \\\\ 1-z = a+b + (-4a-3b)z \\\\ \implies a=-2, b=3[/tex]
[tex]\implies \displaystyle A(z) = \frac16\left(\frac3{1-4z} - \frac2{1-3z}\right)[/tex]
Recall that for |z| < 1, we have
[tex]\displaystyle \sum_{n=0}^\infty z^n = \frac1{1-z}[/tex]
Then if |4z| < 1, we can write A(z) as the sum of two convergent geometric series,
[tex]\displaystyle A(z) = \frac12 \sum_{n=0}^\infty (4z)^n - \frac13 \sum_{n=0}^\infty (3z)^n[/tex]
and some rewriting to put this in the canonical G.F. form lets us easily pick out [tex]a_n[/tex].
[tex]\displaystyle A(z) = \sum_{n=0}^\infty \left(\frac{(4z)^n}2 - \frac{(3z)^n}3 \right) \\\\ A(z) = \sum_{n=0}^\infty \left(\frac{2^{2n}}2 - \frac{3^n}3\right)z^n \\\\ A(z) = \sum_{n=0}^\infty \left(2^{2n-1} - 3^{n-1}\right)z^n[/tex]
[tex]\implies \boxed{a_n = 2^{2n-1} - 3^{n-1}}[/tex]
