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S=1+1+22!+1+2+223!+1+2+22+234!+⋯\large S= 1 + \dfrac{1+2}{2!} + \dfrac{1+2+2^2}{3!} + \dfrac{1+2+2^2+2^3}{4!} + \cdots S=1+2!1+2+3!1+2+22+4!1+2+22+23+⋯
Given that S=Ae2+Be+CS = Ae^2 + Be + C S=Ae2+Be+C, where A,BA,BA,B and CCC are integers, and e=limx→0(1+x)1/x≈2.718\displaystyle e= \lim_{x\to 0} (1+x)^{1/x} \approx 2.718 e=x→0lim(1+x)1/x≈2.718, find the value of A×B×CA\times B \times CA×B×C.
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