Vtyjkyuk

Consider the following linear difference equation
$$ f_k = 1 + frac12 f_k+1 + frac12 f_k-1, 1le k le n-1$$

with $f_0 = f_n = 0$. How do I find the solution? I consider the homogeneous version
$$ f_k+1 - 2f_k + f_k-1 = -1$$

and found that the solution is $C_1 + C_2 k$. But I don't know how to find a particular solution to the non-homogeneous equation.

Easily you can determine that for the homogeneous recurrence equation

$$
f^h_k+1-2f^h_k+f^h_k-1 = 0
$$

the solution is

$$
f^h_k = c_1 + c_2 k
$$

now, a particular solution which should be polynomial, you can propose

$$
f^p_k = c_1 + c_2 k + c_3 k^2
$$

and the coefficient's determination is obtained by substitution in

$$
f^p_k+1-2f^p_k+f^p_k-1 + 1 = 0
$$

obtaining the condition

$$
1+2c_3 = 0Rightarrow c_3 = -frac 12
$$

and finally

$$
f_k = f^h_k + f^p_k = c_1 + c_2 k-frac 12 k^2
$$

Apart from trying what Will Jagy said, you can also note that, for $n=2,3,4,ldots$,
$$beginalignsum_k=2^n,(n-k+1)cdot(-1)&=sum_k=2^n,(n-k+1),left(f_k-2,f_k-1+f_k-2right)
\&=f_n+sum_k=2^n-1,big((n-k+1)-2(n-k)+(n-k-1)big),f_k\
&phantomaaaaa+big((n-2)-2(n-1)big),f(1)+(n-1),f(0)
\
&=f_n-n,f(1)+(n-1),f(0),.
endalign$$
Thus, for every $n=0,1,2,ldots$,
$$f_n(n)=n,f(1)-(n-1),f(0)-sum_k=2^n,(n-k+1)=n,f(1)-(n-1),f(0)-fracn(n-1)2,.$$