Question

The value of the expression $1+\frac{3}{2!}+\frac{7}{3!}+\frac{15}{4!}+.........\infty$ is?

• A. $e(e+1)$
• B. $3e-1$
• C. $e(e-1)$
• D. $3e$

Answer 1

The correct option is C $e(e-1)$

We know that taylor's series of $e^x$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...........$

So, $1+\frac{3}{2!}+\frac{7}{3!}+\frac{15}{4!}+........$

$1+\frac{2^2-1}{2!}+\frac{2^3-1}{3!}+\frac{2^4-1}{4!}+........$

$(1+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+......)-(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+........)$

Adding and subtracting by $2$, we get

$(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+......)-(2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+........)$

$(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+......)-(1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+........)$

Therefore, $(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+......)-(1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+........)=e^2-e=e(e-1)$