Question

If $S=1+\frac{3}{1!}+\frac{5}{2!}+\frac{7}{3!}+...\infty$, then $S$ is equivalent to:

• A. $2e$
• B. $e$
• C. $e-1$
• D. $3e$

Answer 1

Answer: (D)

$3e$

As we know that after applying the Taylor series expansion for $e$, we get

$e=1+\frac{1}{1!}+\frac{1}{2!}+...\infty$

$S=1+\frac{3}{1!}+\frac{5}{2!}+...\infty$

$S-e=\frac{2}{1!}+\frac{4}{2!}+...\infty$

Dividing by $2$ throughout, we get

$\frac{S-e}{2}=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+...\infty =1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\infty =e$

$\Rightarrow S-e=2e$

$\Rightarrow S=3e$