Question

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

Answer 1

Consider the function:
$f(x) = 1 + \frac{(1 + x)}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots$
This represents the series expansion for $e^{1+x}$.

We also have:
$\frac{e^{1+x}}{1+x} = 1 + \frac{(1 + x)}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots$
Identifying the coefficient of $x^2$ in the right-hand side (RHS), we find:
$\text{Coefficient of } x^2 \text{ in RHS: } 1 + \frac{2C_2}{3} + \frac{3C_2}{4} + \dots = a.$
For the left-hand side (LHS), we consider:
$e \left(1 + \frac{x^2}{2!}\right) \left(1 - x + \frac{x^2}{2!}\right)$

This simplifies to terms where the coefficient of $x^2$ matches the expansion on the RHS: $e - e + \frac{e}{2!} = a.$
For the series for $b$, we have:
$b = 1 + \frac{2}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \dots = e^2.$
Finally, we evaluate:
$\frac{2b}{a^2} = 8.$