Let alpha = sum_{n=101}^{200} 2^n sum_{k=101} | vedclass

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(A) Given $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$.Expanding the summation,we have:$\alpha = \frac{1}{101!} (2^{101} + 2^{102} +

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(A) Given $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$.Expanding the summation,we have:$\alpha = \frac{1}{101!} (2^{101} + 2^{102} + \dots + 2^{200}) + \frac{1}{102!} (2^{102} + 2^{103} + \dots + 2^{200}) + \dots + \frac{1}{200!} (2^{200})$.This can be written as:$\alpha = \sum_{n=101}^{200} \frac{1}{n!} \sum_{j=n}^{200} 2^j$.Using the geometric series sum formula $\sum_{j=n}^{200} 2^j = \frac{2^n(2^{201-n}-1)}{2-1} = 2^{201} - 2^n$.Substituting this back into the expression for $\alpha$:$\alpha = \sum_{n=101}^{200} \frac{2^{201} - 2^n}{n!} = b$.Therefore,$\frac{a}{b} = 1$.

Let $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$ and $b = \sum_{n=101}^{200} \frac{2^{201}-2^n}{n !}$. Then,$\frac{a}{b}$ is

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