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The given multivariable recurrence relation is that for every $n \geq 1$ $$S_{n + 1} = T_n - S_n$$ where $S_1 = \dfrac{3}{5}$ and $T_1 = 1$. Both $T_n$ and $S_n$ depend on the following condition $$ \dfrac{T_n}{S_n} = \dfrac{T_{n + 1}}{S_{n + 1}} = \dfrac{T_{n + 2}}{S_{n + 2}} = \dots $$ The goal is to evaluate $$\sum\limits_{n \geq 0} \left(S_{n + 2} + S_{n + 1}\right)^2 (-1)^n$$

Since the change between $T_n$ and $T_{n + 1}$ is not constant, I believe that the way to approach this problem is to have all terms with consistent coefficient. However, I am not skillful enough to simplify the summation into a single variable.

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    $\begingroup$ Love this question +1 $\endgroup$ – mick Feb 7 '17 at 1:07
  • $\begingroup$ Write T in terms of S !! $\endgroup$ – mick Feb 7 '17 at 1:11
  • $\begingroup$ Woo! That, I will admit, was an awesome question. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 1:21
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Notice that

$$\frac53=\frac{T_1}{S_1}=\frac{T_n}{S_n}$$

Thus, $T_n=\frac53S_n$. Putting this in, we get

$$S_{n+1}=\frac23S_n$$

which is a geometric sequence. The general form is then $S_n=\frac35\times\left(\frac23\right)^n$, so we have

$$\text{Sum}=\frac49\sum_{n\ge0}a^n$$

where $a=-\frac49$ is a very simple geometric series.

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    $\begingroup$ I wanted to say that lol. Not so hard like you Said imho :) $\endgroup$ – mick Feb 7 '17 at 1:20
  • $\begingroup$ @mick Lmao, I tried to solve this 2 different ways before this, and then it clicked. XD $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 1:21
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    $\begingroup$ Hehe! I did not see that sweet move coming! I overthought too much about simplifying the summation! Nice method. :) $\endgroup$ – NasuSama Feb 7 '17 at 1:22
  • $\begingroup$ @NasuSama Ha...ha....:P... I too tried and failed many times in the past 10 minutes. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 1:22
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    $\begingroup$ Funny guys here $\endgroup$ – mick Feb 7 '17 at 1:27

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