Solve the following recurrence relation by generating its direct formula: $$a_n = 3a_{n-1} + 2n, a_0 = 1$$ Use the direct formula to find the $10th$ term of the recurrence relation.
My attempt:
$3(10-1) + 2(10)$
$3(9) + 20$
$27 + 20$
$10th$ term = $47$
Is this correct?
On
We have $$a_n=3a_{n-1}+2n\implies a_{n+1}=3a_n+2(n+1)\implies a_{n+2}=3^2a_n+3^1\cdot2(n+1)+2(n+2)$$ so we see that $$a_{n+k}=3^ka_n+2\sum_{i=0}^{k-1}3^i(n+k-i)$$ Hence substituting $n=0, k=10$ gives $$a_{10}=3^{10}a_0+2\sum_{i=0}^93^i(9-i)$$ Can you finish?
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$$a_{n+1}=3a_n+2(n+1) $$ $$a_{n+2}=3a_{n+1}+2 (n+2)$$
thus $$a_{n+2}-4a_{n+1}+3a_n=2$$
the characteristic equation is $$x^2-4x+3=0$$ the roots are $$x_1=1\;\;x_2=3$$ $$a_n=A+B.3^n+\alpha n$$
$$a_0=1 \implies A+B=1$$ $$a_1=5\implies A+3B+\alpha=5$$ $$a_2=19\implies A+9B+2\alpha=19$$
thus $$2B+\alpha=4$$ and $$6B+\alpha=14$$
$$B=\frac {5}{2} \;\;A=-\frac {3}{2}$$ Finally $$B=5/2 \; A=-3/2 \;\;\alpha=-1$$ $$a_n=\frac {5}{2}3^n-\frac {3}{2}-n $$
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The recurrence is $$ a_n = 3a_{n-1} + 2n \tag{1} $$ for $n\geq1$ where $a_0=1$ . If you want to compute using the recurrence relation only (as your post indicates), note that in order to find $a_{10}$ we need the value of $a_9$ and thus the value of $a_8,a_7,\dotsc, a_1$. To begin equation (1) implies that $$ a_1=3a_0+2(1)=3(1)+2(1)=5 $$ and thus using equation (1) again $$ a_2=3a_1+2(2)=3(5)+2(2)=19 $$ Proceed in a similar fashion to compute $a_3,\dotsc, a_9$ and finally $a_{10}$.
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This is a classic example of a problem that can be solved with generating functions. By manipulating the recurrence, you should find that the generating function for the sequence $(a_n)$ is $$f(x) := \sum_{n=1}^{\infty} a_n x^n = \frac{-3 x^3+6 x^2-5 x}{(x-1)^2 (3 x-1)}$$ whose $x^n$ coefficient is not toooo painful to extract.
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For a linear difference equation we break the problem up into $2$ parts: find the general solution to the homogeneous equation and then add any particular solution to the inhomogeneous equation to get the general solution. For the homogeneous equation $$a_{h,n}-3a_{h,n-1}=0$$ Take the general solution to be $a_{h,n}=C\cdot r^n$. Then $$C\cdot r^n-3C\cdot r^{n-1}=C\cdot r^{n-1}(r-3)=0$$ So $r=3$. For the inhomogeneous equatio we hope for a solution of the form $a_{p,n}=An+B$. Then $$a_{p,n}=An+B=3a_{p,n-1}+2n=3\left(A(n-1)+B\right)+2n=3An-3A+3B+2n$$ So $-2An=2n$ therefore $A=-1$ and $3A=-3=2B$ so $B=-\frac32$ and our general solution is $$a_n=a_{h,n}+a_{p,n}=C\cdot3^n-n-\frac32$$ The initial condition is $$a_0=C-\frac32=1$$ So $C=\frac52$ and the solution to the initial value problem is $$a_n=\frac52\cdot3^n-n-\frac32$$ This agrees with the first few iterates of @FoobazJohn.
Divide by $3^n$ and define $u_n=a_n/3^n$. Solve the resulting equation for $u_n$ (this involves computing a sum) and retrieve $a_n$.