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Consider the following infinite series. Complete parts (a) through (c). a. Find the first four partial sums \( \mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3} \), and \( \mathrm{S}_{4} \). \( \mathrm{S}_{1}=\square \) (Simplify your answer. Type integers or decimals.) b. Find a formula for the nth partial sum \( \mathrm{S}_{\mathrm{n}} \) of the infinite series. \( \mathrm{S}_{\mathrm{n}}=\square \) (Use integers or fractions for any numbers in the expression.) Use the formula to find the next four partial sums \( \mathrm{S}_{5}, \mathrm{~S}_{6}, \mathrm{~S}_{7} \), and \( \mathrm{S}_{8} \) of the infinite series. \( \mathrm{S}_{5}=\square \) (Simplify your answer. Type integers or decimals.) c. Make a conjecture for the value of the series. \( \mathrm{Sel}^{k} \). answer box to complete your choice. A. The correct choice below and, if necessary, fill in the A. The value of the infinite series is (Type an integer or a simplified fraction.) B. The infinite series diverges.

Ask by Lambert Vega. in the United States
Mar 15,2025

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**Part (a):** - \( S_1 = 1 \) - \( S_2 = 1.25 \) - \( S_3 \approx 1.3611 \) - \( S_4 \approx 1.4236 \) **Part (b):** \[ S_n = \sum_{k=1}^{n} \frac{1}{k^2} \] - \( S_5 \approx 1.4636 \) - \( S_6 \approx 1.4914 \) - \( S_7 \approx 1.5118 \) - \( S_8 \approx 1.5274 \) **Part (c):** The infinite series converges to \( \frac{\pi^2}{6} \approx 1.64493 \).

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Beyond the Answer

The first four partial sums of an infinite series can often provide great insight into the behavior of the series itself. Partial sums are simply the sums of the first \( n \) terms of a sequence, and they can exhibit interesting patterns, often giving clues about convergence or divergence. For example, if the partial sums seem to approach a specific number as \( n \) increases, it hints that the series might converge to that value. In practical applications, understanding partial sums can help in fields like physics, engineering, or finance where series often find a place in modeling phenomena. Whether it's predicting the trajectory of a rocket or managing the growth of investments, these sums help encapsulate infinite processes into manageable finite calculations, drawing us closer to true values in a quantifiable way.

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\[ y=x^{3}, \quad 0 \leq x \leq 2 \] Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interv \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note t as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius and the radical measures the arc length that is the width of a band, \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying, \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 14 \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
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