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The recurrence is $K(n)=2K(n-1)-K(n-2)+C$ where $C$ is a constant.
What I have tried is substituting $2K(n-1)$ as we do in fibonnacical recurrences. It didn't gave me a fruitful expression!
Can someone help in solving it?
Not a homework problem.

$$
begin{bmatrix}
K(n+1) \ K(n)
end{bmatrix}=A
begin{bmatrix}
K(n) \ K(n-1)
end{bmatrix}+
begin{bmatrix}
C \ 0
end{bmatrix}
$$

where

$$A=
begin{bmatrix}
2 & -1 \ 1 & 0
end{bmatrix}.
$$

Therefore,

$$
begin{bmatrix}
K(n+1) \ K(n)
end{bmatrix}=A^n
begin{bmatrix}
K(1) \ K(0)
end{bmatrix}+
left(sum_{k=1}^{n-1}A^k+Iright)
begin{bmatrix}
C \ 0
end{bmatrix}
$$

and

$$
K(n)=nK(1)-(n-1)K(0)+frac{1}{2}Cn(n-1)
$$

by noticing that

$$
A^k=
begin{bmatrix}
k+1 & -k \ k & k-1
end{bmatrix}.
$$

The other answers are way too complicated for this particular problem.

They're useful in more general cases, but they're completely overkill here.

begin{align*}
K(n) &= 2 K(n - 1) - K(n - 2) + C \
K(n) - K(n - 1) &= K(n - 1) - K(n - 2) + C
end{align*}

Just look at this equation for a few seconds.

It's literally telling you that the difference between successive elements increases by $C$ every step.

So... just go ahead and count how many times you add $C$ to the difference $K(1) - K(0)$:

$$K(n) = K(0) + sum_{k=1}^{n} K(1) - K(0) + (k - 1) C$$

Notice you don't need any linear algebra, eigenvectors, or other higher-level math for this problem. It's just algebraic manipulation.

I'll leave the last step of simplifying the summation to you.

Edit:

or I'll just do it for you myself, since you seem to think it leads to another recurrence...

begin{align*}
K(n) &= K(0) + n(K(1) - K(0)) + Csum_{k=1}^{n} (k-1) \
&= K(0) + n(K(1) - K(0)) + C frac{n(n-1)}{2}
end{align*}

To keep things general, suppose $k_0=A$ and $k_1=B$. Then the next few terms are:
$$k_2=2A-B+C\
k_3=4A-3B+3C\
k_4=6A-5B+6C\
k_5=8A-7B+10C\
k_6=10A-9B+15C$$

The pattern seems to be (for $nge2$) that 2 more $A$'s are added, 2 more $B$'s are subtracted, and $n-1$ more $C$'s are added,
$$k_n=(2n-2)A-(2n-3)B+frac{(n-1)(n)}{2}C$$

A proof by strong induction along with some messy algebra will give you your answer

More standard way. Rewrite your equation:
$$
K(n)-2K(n-1)+K(n-2)=C tag{1}label{1}
$$
Solution of this is $K=K_0+K_{part}$, where
$$
K_0(n)-2K_0(n-1)+K_0(n-2)=0tag{2}label{2}
$$
and $K_{part}$ is any solution of $eqref{1}$.

Now we're finding solution of homogenuous equation $eqref{2}$ in a form $K(n)=mathrm{const}cdot lambda^n$ and get
$$lambda^{n}-2lambda^{n-1}+lambda^{n-2}=0Longrightarrow lambda^2-2lambda+1=0;$$
$lambda_1=lambda_2=1$, and
$$
K_0(n)=A+Bn.tag{3}label{3}
$$

$K_{part}$ we'll find in a form $K_{part}=alpha n^2$ (polynom of degree $0$ and $1$ we used yet). Substitute it in $eqref{1}$ and get $2alpha=C$.

So, solution is
$$
K(n)=A+Bn+frac{Cn^2}{2}.
$$

If you prefer, we can take $K(0)=A$ and $K(1)=A+B+C/2$; hence,
$$
K(n)=K_0+(K_1-K_0)n+frac{Cn(n-1)}{2}
$$