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Question

If  $$\displaystyle \frac{e - 1}{e + 1} = \dfrac{\dfrac{1}{a!} + \dfrac{1}{b!} + \dfrac{1}{6!} + ...}{ \dfrac{1}{1!} + \dfrac{1}{c!} + \dfrac{1}{d!} + ...}$$ then find $$ a+b -c +d$$


Solution

We know,  $$\displaystyle e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...........$$
$$\therefore \displaystyle e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...........(i)$$
and $$\therefore \displaystyle e^{-1} = 1-1+\frac{1}{2!}-\frac{1}{3!}+...........(ii)$$
Now  $$\displaystyle \frac{e-1}{e+1}=\frac{e-1}{e+1}.\frac{e-1}{e-1}=\frac{e^2+1-2e}{e^2-1}$$
$$\displaystyle \quad =\frac{e+e^{-1}-2}{e-e^{-1}}=\frac{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+........}{\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+........}$$, using $$(i)$$ and $$(ii)$$

Mathematics

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