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Consider Following Expression:

$S(k)=\frac{a_1 (-1)^k}{k!}+a_2 \sum _{l_1=0}^k \frac{(-1)^k}{l_1! \left(k-l_1\right)!}+a_3 \sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \frac{(-1)^k}{l_2! \left(l_1-l_2\right)! \left(k-l_1\right)!}+a_4 \sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \sum _{l_3=0}^{l_2} \frac{(-1)^k}{l_3! \left(l_2-l_3\right)! \left(l_1-l_2\right)! \left(k-l_1\right)!}+a_5 \sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \sum _{l_3=0}^{l_2} \sum _{l_4=0}^{l_3} \frac{(-1)^k}{l_4! \left(l_3-l_2\right)! \left(l_2-l_3\right)! \left(l_1-l_2\right)! \left(k-l_1\right)!}+\text{...}$

I want to Express these in a closed form.like this:

$S(k)=\sum _{i=1}^n (\text{?!})a_i$

any ideas how could it be?

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2 Answers 2

Use the binomial formula $$ \sum_{l=0}^k \binom{k}{l}=2^k, $$ which can be rewritten as $$ \sum_{l=0}^k\frac{(-1)^k}{l!(k-l)!}=\frac{(-2)^k}{k!}. $$

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As answered by LutzL, continuing the summations using the binomial formula, you should arrive to $$S_1(k)= \frac{(-1)^k}{k!}$$ $$S_2(k)= \sum _{l_1=0}^k \frac{(-1)^k}{l_1! \left(k-l_1\right)!}=\frac{(-2)^k}{k!}$$ $$S_3(k)=\sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \frac{(-1)^k}{l_2! \left(l_1-l_2\right)! \left(k-l_1\right)!}=\frac{(-3)^k}{k!}$$

$$S_4(k)= \sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \sum _{l_3=0}^{l_2} \frac{(-1)^k}{l_3! \left(l_2-l_3\right)! \left(l_1-l_2\right)! \left(k-l_1\right)!}=\frac{(-4)^k}{k!}$$

$$S_5(k)=\sum _{l_1=0}^k \sum _{l_2=0}^{l_1} \sum _{l_3=0}^{l_2} \sum _{l_4=0}^{l_3} \frac{(-1)^k}{l_4! \left(l_3-l_2\right)! \left(l_2-l_3\right)! \left(l_1-l_2\right)! \left(k-l_1\right)!}=\frac{(-5)^k}{k!}$$ which clearly show the pattern and, finally, $$S(k)=\sum _{i=1}^n a_i S_i(k)=???$$

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