The lowest integer which is greater than left | vedclass

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Question

(A) Let $P = \left(1 + \frac{1}{x}\right)^x$ where $x = 10^{100}$.Using the binomial expansion for $(1 + \frac{1}{x})^x$:$P = 1 + x \cdot \frac{1}{x}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let $P = \left(1 + \frac{1}{x}\right)^x$ where $x = 10^{100}$.Using the binomial expansion for $(1 + \frac{1}{x})^x$:$P = 1 + x \cdot \frac{1}{x} + \frac{x(x-1)}{2!} \cdot \frac{1}{x^2} + \frac{x(x-1)(x-2)}{3!} \cdot \frac{1}{x^3} + \dots$$P = 1 + 1 + \frac{1}{2!}(1 - \frac{1}{x}) + \frac{1}{3!}(1 - \frac{1}{x})(1 - \frac{2}{x}) + \dots$Since $x = 10^{100}$ is very large,each term $(1 - \frac{k}{x})$ is slightly less than $1$.Thus,$P We know that $e \approx 2.718$.Also,$P > 1 + 1 = 2$.Therefore,$2 The lowest integer greater than $P$ is $3$.

The lowest integer which is greater than $\left(1+\frac{1}{10^{100}}\right)^{10^{100}}$ is $.....$

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