Vtyjkyuk

$$24a(n)=26a(n-1)-9a(n-2)+a(n-3)$$
$$a(0)=46, a(1)=8, a(2)=1$$
$$sumlimits_k=3^inftya(k)=2^-55$$
How can I prove it?

$$24r^3=26r^2-9r+1$$

solutions
$$r=frac12,frac13,frac14$$

$$a_n=xleft(frac12right)^n+yleft(frac13right)^n+zleft(frac14right)^n$$
Determine $x,y,z$ using base conditions

$$a_0=46=x+y+z$$
$$a_1=8=x/2+y/3+z/4$$
$$a_2=1=x/4+y/9+z/16$$

$$fbox
x=4, y=-54,z=96
$$

$$a_n=4left(frac12right)^n-54left(frac13right)^n+96left(frac14right)^n$$

$$sumlimits_k=3^inftya(k)=sumlimits_k=3^inftyleft(4left(frac12right)^k-54left(frac13right)^k+96left(frac14right)^kright)$$

$$sum_k=3^infty4left(frac12right)^k=4sum_k=3^inftyleft(frac12right)^k=4cdotfracleft(frac12right)^31-frac12=1$$

$$54sum_k=3^inftyleft(frac13right)^k=54fracleft(frac13right)^31-frac13=3$$
$$96sum_k=3^inftyleft(frac14right)^k=96cdotfracleft(frac14right)^31-frac14=2$$

$$sumlimits_k=3^inftya_k=1-3+2=0$$

Let $S=sumlimits_k=3^inftya(k)$. Then
$$
24S = 26(S+a(2))-9(S+a(1)+a(2))+(S+a(0)+a(1)+a(2))
$$
This gives $S=0$.

Consider the generating function $$f(x)=a_0x^0+a_1x^1+a_2x^2+cdots+a_nx^n+cdots.tag1$$

We have $$f(x)cdot x^3=a_0x^3+a_1x^4+a_2x^5+cdots+a_nx^n+3+cdots;tag2$$
$$f(x)cdot x^2=a_0x^2+a_1x^3+a_2x^4+cdots+a_nx^n+2+cdots;tag3$$
$$f(x)cdot x=a_0x^1+a_1x^2+a_2x^3+cdots+a_nx^n+1+cdots.tag4$$
Thus, by $(2)-9times(3)+26times(4)-(1)$, we obtain
beginalign*
f(x)(x^3-9x^2+26x-1)&=-9a_0x^2+26a_0x^1+26a_1x^2-a_0x^0-a_1x^1-a_2x^2\
&=-207x^2+1188x-46
endalign*

Hence $$boxedf(x)=frac-207x^2+1188x-46x^3-9x^2+26x-1.$$

Thus $$sum_k=0^infty a_k=f(1)=55.$$

As a result, $$sum_k=3^infty a_k=sum_k=0^infty a_k-sum_k=0^2 a_k=55-55=0.$$