The approximate value of (1.0002)^{3000} is | vedclass

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(A) Using the binomial expansion $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ Given $(1.0002)^{3000} = (1 + 0.0002)^{3000}$. Here $n = 3000$ an

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$1.6$

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(A) Using the binomial expansion $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ Given $(1.0002)^{3000} = (1 + 0.0002)^{3000}$. Here $n = 3000$ and $x = 0.0002$. Using the first two terms of the expansion: $(1 + 0.0002)^{3000} \approx 1 + (3000)(0.0002) + \frac{3000 	imes 2999}{2} 	imes (0.0002)^2$. Since $(0.0002)^2 = 0.00000004$,the higher-order terms are negligible. Therefore,$(1.0002)^{3000} \approx 1 + 0.6 = 1.6$.

The approximate value of $(1.0002)^{3000}$ is

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