Question

The sum of $\frac{2}{1!} + \frac{6}{2!} + \frac{12}{3!} + \frac{20}{4!} +$.......is.

• A. $\frac{3e}{2}$
• B. $e$
• C. $2e$
• D. $3e$

Answer 1

Answer: (D)

$3e$

Answer 2

Let $x^{- 4}$ and let

$- 1$

$\frac{1}{2.3} + \frac{1}{4.5} + \frac{1}{6.7} + ... =$

$\log(2/e)$

$\log(e/2)$

⇒ $b = a - \frac{a^{2}}{2} + \frac{a^{3}}{3} - \frac{a^{4}}{4} + ..$=$b + \frac{b^{2}}{2!} + \frac{b^{3}}{3!} + \frac{b^{4}}{4!} + ...\infty =$= $\log_{e}a$

∴ n^th term of given series

$\log_{e}b$ or $a$

Now, sum = $e^{a}$.