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The best part of this solution is that it's deceptively straight forward. If somebody were to see all of these "fancy" math symbols and such they would instinctively feel intimidated, but if you really look at what you're doing here it's really just very smart equation manipulation.

$\sum_{n = 1}^{\infty} \frac {1}{1 - x^n} - \frac{1}{1 - x^{n + 1}}$. So, when $n = 1$ then you have $\frac {1}{1 - x} - \frac {1}{1 - x^2}$. Add $n = 2$ you have $\frac {1}{1 - x^2} - \frac {1}{1 - x^3}$. Notice how the first term of this cancels with the second term of the first. This is called a telescopic series. Thus, you will only have the first term ( $\frac{1}{1 - x}$) and the last term, $\lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0}$.

When solving this question, I first did the manipulations to the summation. I then replaced $2/3$ with $x$ to see if there was some clever power series manipulations that could be done. As it turned out, it was just a telescoping sum, and replacing $2/3$ with $x$ was not really necessary. However, $x$ is easier to type repeatedly than $2/3$, and so, I showed the generalization.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestIf I'm not mistaken, this was a Putnam problem!

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Let's try to generalize. For $|x| < 1$ we have the following:

$S(x) := \displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n}{(1-x^n)(1-x^{n+1})}$ $= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n(1-x)}{(1-x^n)(1-x^{n+1})}$

$= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n - x^{n+1}}{(1-x^n)(1-x^{n+1})}$ $= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{(1- x^{n+1}) - (1-x^n)}{(1-x^n)(1-x^{n+1})}$

$= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty}\left[\dfrac{1}{1-x^n} - \dfrac{1}{1-x^{n+1}}\right]$ $= \dfrac{1}{1-x}\left[ \dfrac{1}{1-x^1} - \dfrac{1}{1-0}\right]$ $= \dfrac{x}{(1-x)^2}$.

Now to tackle the sum in question:

$\displaystyle\sum_{n = 1}^{\infty} \dfrac{6^n}{(3^n-2^n)(3^{n+1}-2^{n+1})}$ $= \displaystyle\sum_{n = 1}^{\infty} \dfrac{3^{2n}\left(2/3\right)^n}{3^{2n+1}\left(1-\left(2/3\right)^n\right)\left(1-\left(2/3\right)^{n+1}\right)}$

$= \dfrac{1}{3}\displaystyle\sum_{n = 1}^{\infty} \dfrac{\left(2/3\right)^n}{\left(1-\left(2/3\right)^n\right)\left(1-\left(2/3\right)^{n+1}\right)}$ $= \dfrac{1}{3}S\left(\dfrac{2}{3}\right)$ $= \dfrac{1}{3} \dfrac{2/3}{\left(1-2/3\right)^2} = \boxed{2}$

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that is incredible genius imo, you should be proud xD

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Forgive my ignorance but what is S(x)? Is it the same as f(x)?

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$S(x)$ is a function of $x$ that I defined. I could have also named it $f(x)$, or $\xi(x)$, or even $\clubsuit(x)$.

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The best part of this solution is that it's deceptively straight forward. If somebody were to see all of these "fancy" math symbols and such they would instinctively feel intimidated, but if you really look at what you're doing here it's really just very smart equation manipulation.

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How will we get

$\sum_{n=1}^{\infty} \Bigg[ -\frac{1}{1-x^{n+1}} \Bigg] = -\frac{1}{1-0}$ ?

Please help , Thank You

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Look at the whole expression. It states

$\sum_{n = 1}^{\infty} \frac {1}{1 - x^n} - \frac{1}{1 - x^{n + 1}}$. So, when $n = 1$ then you have $\frac {1}{1 - x} - \frac {1}{1 - x^2}$. Add $n = 2$ you have $\frac {1}{1 - x^2} - \frac {1}{1 - x^3}$. Notice how the first term of this cancels with the second term of the first. This is called a telescopic series. Thus, you will only have the first term ( $\frac{1}{1 - x}$) and the last term, $\lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0}$.

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Good explanation too !

Thank You !

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Wow! Just great, Jimmy!

How did you even think of S(x)?

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When solving this question, I first did the manipulations to the summation. I then replaced $2/3$ with $x$ to see if there was some clever power series manipulations that could be done. As it turned out, it was just a telescoping sum, and replacing $2/3$ with $x$ was not really necessary. However, $x$ is easier to type repeatedly than $2/3$, and so, I showed the generalization.

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