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(B) To find the sum $S_n = \sum_{k=1}^{n} \sqrt{k}$ for large $n$,we can use the definition of a definite integral as the limit of a Riemann sum. We w

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$\frac{2}{3} n^{\frac{3}{2}}$

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(B) To find the sum $S_n = \sum_{k=1}^{n} \sqrt{k}$ for large $n$,we can use the definition of a definite integral as the limit of a Riemann sum. We write the sum as: $S_n = n \sqrt{n} \left( \frac{1}{n} \sum_{k=1}^{n} \sqrt{\frac{k}{n}} ight)$ As $n 	o \infty$,the term in the parenthesis approaches the definite integral: $\lim_{n 	o \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{\frac{k}{n}} = \int_{0}^{1} \sqrt{x} \, dx$ Evaluating the integral: $\int_{0}^{1} x^{\frac{1}{2}} \, dx = \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} ight]_{0}^{1} = \frac{2}{3}$ Therefore,for large $n$: $S_n \approx n \sqrt{n} \cdot \frac{2}{3} = \frac{2}{3} n^{\frac{3}{2}}$

For a sufficiently large value of $n$,the sum of the square roots of the first $n$ positive integers,i.e.,$\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n}$,is approximately equal to:

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