Let a = 1 + C 2 2 3 ! + C 2 3 4 ! + C 2 4 5 ! + … , b = 1 + C 0 1 + 1 C 1 1 ! + C 0 2 + 2 C 1 + 2 C 2 2 ! + C 0 3 + 3 C 1 + 3 C 2 + 3 C 3 3 ! + … Then 2 b a 2 is equal to ______ . [2024]

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Solution

Q.
Let $a = 1 + \frac{{}^{2}{C_{2}}}{3!} + \frac{{}^{3}{C_{2}}}{4!} + \frac{{}^{4}{C_{2}}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}{C_{0}} +^{1} C_{1}}{1!} + \frac{{}^{2}{C_{0}} +^{2} C_{1} +^{2} C_{2}}{2!} + \frac{{}^{3}{C_{0}} +^{3} C_{1} +^{3} C_{2} +^{3} C_{3}}{3!} + \dots$ Then $\frac{2 b}{a^{2}}$ is equal to ______ . [2024]

Ans.
(8)

Given, $a = 1 + \frac{{}^{2}C_{2}}{3!} + \frac{{}^{3}C_{2}}{4!} + \frac{{}^{4}C_{2}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}C_{0} + {}^{1}C_{1}}{1!} + \frac{{}^{2}C_{0} + {}^{2}C_{1} + {}^{2}C_{2}}{2!} + . . .,$

$\Rightarrow b = 1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + \dots = e^{2}$ $[∵ e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots]$

Now, $a = 1 + \sum_{r = 2}^{\infty} \frac{{}^{r}C_{2}}{(r + 1)!} = 1 + \sum_{r = 2}^{\infty} \frac{r (r - 1)}{2 (r + 1)!}$

$= 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{(r + 1) r - 2 r}{(r - 1)!} = 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{1}{(r - 1)!} - \frac{1}{2} \sum_{r = 2}^{\infty} \frac{2 r}{(r + 1)!}$

$= 1 + \frac{1}{2} (\frac{1}{1!} + \frac{1}{2!} + . . .) - \sum_{r = 2}^{\infty} \frac{(r + 1) - 1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - \sum_{r = 2}^{\infty} \frac{1}{r!} + \sum_{r = 2}^{\infty} \frac{1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - (e - \frac{1}{1!} - \frac{1}{0!}) + (e - \frac{1}{1!} - \frac{1}{0!} - \frac{1}{2!})$

$= 1 + \frac{e}{2} - \frac{1}{2} - e + 2 + e - 2 - \frac{1}{2} = \frac{e}{2} \Rightarrow \frac{2 b}{a^{2}} = \frac{2 e^{2}}{\frac{e^{2}}{4}} = 8$

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