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More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\alpha^p = F_{\!p}\,\alpha^q,$$ where $\alpha = (1+\sqrt5)/2$, and the analogous formula with $\alpha$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}.\tag{2} $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\alpha^p = F_{\!p}\,\alpha^q,$$ where $\alpha = (1+\sqrt5)/2$, and the analogous formula with $\alpha$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}.\tag{2} $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Similar identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\alpha^p = F_{\!p}\,\alpha^q,$$ where $\alpha = (1+\sqrt5)/2$, and the analogous formula with $\alpha$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}.\tag{2} $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.