Hello Saffjdaf,
1.) We are given the homogeneous recursion:
$$a_{n}=-a_{n-1}$$ where $$a_1=1$$
The idea behind solving homogeneous difference equations is to assume a solution of the form:
$$a_n=c_1r^n$$ where $$c_1,r\ne0$$
And so, we may write the given difference equation as:
$$c_1r^n+c_1r^{n-1}=0$$
Dividing through by $$c_1r^{n-1}$$ we obtain the characteristic equation:
$$r+1=0\implies r=-1$$
And so our solution must be:
$$a_n=c_1(-1)^n$$
Now, we may use the given initial value to determine the parameter $c_1$:
$$a_1=c_1(-1)^1=-c_1=1\implies c_1=-1$$
And so we find the closed form for the solution is:
$$a_{n}=(-1)^{n+1}$$
Thus, the sequence oscillates from 1 to -1 back to 1 indefinitely. Terms with an odd subscript are 1 and terms with an even subscript are -1.
2.) The only difference between this problem as the first is the initial value, we we know:
$$a_n=c_1(-1)^n$$
Next, using the initial value, we find:
$$a_1=c_1(-1)^1=-c_1=2\implies c_1=-2$$
And so we have:
$$a_{n}=2(-1)^{n+1}$$
In fact, if we simply use $a_1$ as the value of the first term, we find:
$$a_1=c_1(-1)^1=-c_1\implies c_1=-a_1$$
and we have:
$$a_{n}=a_1(-1)^{n+1}$$
Thus the sequence oscillate from $a_1$ to $-a_1$ and back again to $a_1$ indefinitely.
3.) This time we have the inhomogeneous recurrence:
$$a_{n}=n+3a_{n-1}$$ where $$a_1=-2$$
Now, we want first to find the solution to the corresponding homogenous difference equation, which has an associated characteristic equation of:
$$r-3=0\implies r=3$$
and so the homogeneous solution is:
$$h_n=c_1(3)^n$$
Next, we assume a particular solution of the form:
$$p_n=An+B$$ where $A$ and $B$ are constants to be determined.
We assume a linear form for the particular solution because the inhomogeneous term is linear and no part of the homogeneous solution is part of this assume linear solution, that is, they are linearly independent.
So, substituting the particular solution into the difference equation, we find:
$$(An+B)-3(A(n-1)+B)=n$$
$$(-2A)n+(3A-2B)=(1)n+(0)$$
Equating coefficients, we find:
$$-2A=1\implies A=-\frac{1}{2}$$
$$3A-2B=0\implies B=\frac{3}{2}A=-\frac{3}{4}$$
And so our particular solution is:
$$p_n=-\frac{1}{n}n-\frac{3}{4}=-\frac{2n+3}{4}$$
Hence, by superposition, we find:
$$a_n=h_n+p_n=c_1(3)^n-\frac{2n+3}{4}$$
Next, we mat use the given initial value to determine $c_1$:
$$a_1=c_1(3)^1-\frac{2(1)+3}{4}=3c_1-\frac{5}{4}=-2\,\therefore\,c_1=-\frac{1}{4}$$
And so, the closed form satisfying all given conditions is:
$$a_n=-\frac{3^n+2n+3}{4}$$
And so we have the next 5 terms:
[TABLE="class: grid, width: 250, align: left"]
[TR]
[TD]$n$[/TD]
[TD]$a_n$[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]-2[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]-4[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]-9[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]-23[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]-64[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]-186[/TD]
[/TR]
[/TABLE]
