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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!

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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 !

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Hint: There are 2 geometric progressions.

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

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