If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$ 
$(1)\;\;$ If $f(n+1)-f(n) = 3$ and $f(0) = 1\;,$ Then $f(n) = \;,$ Where $n\in \mathbb{W}$
$(2)\;\;$ If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$
Where $n\in \mathbb{W}$

$\bf{My\; Try\;, (1)::}$ Let $f(n)=a_{n}\;,$ Then Our functional equation convert into $a_{n+1}-a_{n} = 3$ and $a_{0}=1$
So $$a_{n+1}=3+a_{n}\;,$$ Now Replcae $n\rightarrow (n-1)\;,$ We get
$$a_{n}=3+a_{n-1}.............(1)$$
Now in a  Similar way, We get $$a_{n}=3+3+a_{n-2}$$
So in a Similar way, We get $$a_{n}=\underbrace{3+3+3+.........+3}_{\bf{n\;  times}}+a_{n-n}=3n+a_{0}$$
So we get $$a_{n}=1+3n\;,$$ Where $n\in \mathbb{W}$
$\bf{(2)}$ Here our equation Convert into $a_{n+2}-5a_{n+1}+6a_{n}=0\;\;,$ and $a_{0}=2\;\;a_{1}=5$
Now I did not Understand How can I solve it, Plz explain me in detail,
Thanks

Let the generating function of the sequence $a$ defined by
$$a_{n + 2} - 5a_{n + 1} + 6a_n= 0$$
be 
$$g(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots$$
Then 
$$-5xg(x) = -5a_0x -5a_1x^2 - 5a_2x^3 - 5a_3x^4 - \dots$$
and
$$6x^2g(x) = 6a_0x^2 + 6a_1x^3 + 6a_2x^4 + 6a_3x^5 + \dots$$
Adding all three up,
$$\begin{align}g(x) - 5xg(x) + 6x^2g(x) &= a_0 + (a_1 - 5a_0)x \\
&+(a_2 - 5a_1 + 6a_0)x^2\\
&+(a_3 - 5a_2 + 6a_1)x^3\\
&+\dots\\
&=2 -5x \end{align}$$
so that
$$(1 - 5x + 6x^2)g(x) = 2 - 5x$$
$$g(x) = \frac{2 - 5x}{(3x - 1)(2x - 1)}$$
It should be relatively more trivial to decompose this into partial fractions. You can then use geometric series and some algebra to determine the general term for the coefficient of $x^n$ in $g(x)$. It will correspond to $a_n$.
