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

Correct Answer: 8

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: 
\[ \frac{2b}{a^2} = 8. \]