b = 1 + frac{{}^1 C_0 + {}^1 C_1}{1!} + frac{ | vedclass

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(B) We know that $\sum_{r=0}^{n} {}^n C_r = 2^n$. Thus,$b = 1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots = \sum_{n=0}^{\infty} \frac{

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$8$

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(B) We know that $\sum_{r=0}^{n} {}^n C_r = 2^n$. Thus,$b = 1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots = \sum_{n=0}^{\infty} \frac{2^n}{n!} = e^2$. Now,$a = 1 + \sum_{n=2}^{\infty} \frac{{}^n C_2}{(n+1)!}$. Using ${}^n C_2 = \frac{n(n-1)}{2}$,we have $a = 1 + \sum_{n=2}^{\infty} \frac{n(n-1)}{2(n+1)!} = 1 + \frac{1}{2} \sum_{n=2}^{\infty} \frac{n(n-1)}{(n+1)!}$. Since $\frac{n(n-1)}{(n+1)!} = \frac{(n+1-1)(n-1)}{(n+1)!} = \frac{(n+1)(n-1) - (n-1)}{(n+1)!} = \frac{n-1}{n!} - \frac{n-1}{(n+1)!} = \left(\frac{n}{(n)!} - \frac{1}{n!}\right) - \left(\frac{n+1}{(n+1)!} - \frac{2}{(n+1)!}\right) = \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right) - \left(\frac{1}{n!} - \frac{2}{(n+1)!}\right)$. Summing this series leads to $a = e/2$. Therefore,$\frac{2b}{a^2} = \frac{2(e^2)}{(e/2)^2} = \frac{2e^2}{e^2/4} = 8$.

$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$
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:

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$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$ 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: