Roots of equation form infinite sequence

The sequence $$a_n$$ has the property that $$a_n$$ and $$a_{n+1}$$ are the roots of the equation $$x^2-c_nx+\frac{1}{3^n}=0$$ and $$a_1=2$$. What is $$\sum_{n=1}^{\infty}c_n?$$

By Vieta's, $$a_{n+1}=\frac{1}{3^na_n}$$ and $$c_n=a_n+a_{n+1}$$. Additionally listing the first numbers in $$a_n$$ $$2,\frac{1}{6},\frac{2}{3}, \frac{1}{18}, \frac{2}{9},\cdots$$ doesn't reveal anything (even though $$c_n$$ seems like a geometric sequence). Thanks!

Instead of simplifying all the terms, try to find a relation between every term in terms of $$a_1$$.

Now, your sequence of $$a$$ shall look like this $$a_1 ,\frac{1}{3a_1},\frac{a_1}{3},\frac{1}{9a_1}\cdots$$

Notice that, every even terms of the sequence are in a $$g.p$$ with common ratio $$\frac{1}{3}$$

And every odd terms are also in $$g.p$$ with common ratio $$\frac{1}{3}$$ again !

We need to find $$\Sigma_{n=1}^{\infty}c_{n} = c_1 + c_2 + c_3 \cdots = (a_1+a_2)+(a_2+a_3) + (a_3+a_4) \cdots$$

Notice that, except the first term i.e $$a_1$$ every other term occurs 2 times, so our sum is simply, $$a_1 + 2(a_3+a_5+a_7+\cdots)+2(a_2+a_4+a_6+\cdots)$$

We know that the 2 sequences in the above expression have a common ratio $$\lt 1$$, So their infinite sum will converge, and hence our answer is $$a_1 + \frac{2a_3}{1-\frac{1}{3}}+\frac{2a_2}{1-\frac{1}{3}} = 2 + \frac{2\cdot\frac{2}{3}}{1-\frac{1}{3}} + \frac{2\cdot\frac{1}{6}}{1-\frac{1}{3}} = 2 + 2 + \frac{1}{2} = \frac{9}{2}$$

So our answer is $$\frac{9}{2}$$. Hope this helps !

Hint: There are 2 geometric progressions.

Show that $$\frac{a_{n+2}}{a_n} = \frac{1}{3}$$.
Hence $$\sum a_{2i+1} = ?? , \sum a_{2i} = ??, \sum c_i = ??$$