The value of $\frac{2\frac{1}{2}}{1!} + \frac{3\frac{1}{2}}{2!} + \frac{4\frac{1}{2}}{3!} + \frac{5\frac{1}{2}}{4!} + \dots \infty$ is

$\sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$ is equal to

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

$1 + \frac{a - bx}{1!} + \frac{(a - bx)^2}{2!} + \frac{(a - bx)^3}{3!} + \dots \infty = $

$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$
Let $a = 1 + \frac{{}^2 C_2}{3!} + \frac{{}^3 C_2}{4!} + \frac{{}^4 C_2}{5!} + \ldots$. Then $\frac{2b}{a^2}$ is equal to:

The expression $\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \end{aligned}$ for $a>0, x \in R$ is equal to: