It is a known fact that $$\sum\limits_{n \geq 0} F_{n}t^{n} = \frac{t}{1-t-t^2}.$$
Now if we want the odd terms just $$\sum\limits_{n \geq 0} F_{2n+1}t^{2n+1} = \frac{1}{2}(F(t)-F(-t)) = \frac{t(1-t^2)}{1-3t^2-t^4}.$$ What I don't get how re-writing the term we end up with $$\sum\limits_{j \geq 2}F_{2j-3}t^j = \sum\limits_{n \geq 0} F_{2n+1}t^{n+2} = \frac{t^2(1-t)}{1-3t-t^2}$$
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Any help or hint would be appreciated.
If $$\sum_{n = 0}^\infty F_{2n+1} t^{2n+1} = \frac{t(1-t^2)}{1 - 3t^2 + t^4}, \tag{1}$$ then dividing both sides by $t$ gives $$\sum_{n=0}^\infty F_{2n+1} (t^2)^n = \frac{1 - t^2}{1 - 3t^2 + (t^2)^2}, \tag{2}$$ hence letting $t^2 = u$ implies $$\sum_{n=0}^\infty F_{2n+1} u^n = \frac{1 - u}{1 - 3u + u^2}. \tag{3}$$ All that remains is to now multiply by $u^2$ and rename $u$ to $t$: $$\sum_{n=0}^\infty F_{2n+1} u^{n+2} = \frac{u^2(1-u)}{1 - 3u + u^2}, \tag{4}$$ therefore $$\sum_{n=0}^\infty F_{2n+1} t^{n+2} = \frac{t^2(1-t)}{1 - 3t + t^2}. \tag{5}$$