Let a_{n} = int_{-1}^{n} left(1 + frac{x}{2} | vedclass

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(C) Given $a_{n} = \int_{-1}^{n} \left(\sum_{k=1}^{n} \frac{x^{k-1}}{k}\right) dx$. Integrating term by term,we get: $a_{n} = \left[ x + \frac{x^{2}}{

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$5$

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(C) Given $a_{n} = \int_{-1}^{n} \left(\sum_{k=1}^{n} \frac{x^{k-1}}{k}\right) dx$. Integrating term by term,we get: $a_{n} = \left[ x + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{3^{2}} + \ldots + \frac{x^{n}}{n^{2}} \right]_{-1}^{n}$. Evaluating at the limits: $a_{n} = \left(n + \frac{n^{2}}{2^{2}} + \frac{n^{3}}{3^{2}} + \ldots + \frac{n^{n}}{n^{2}}\right) - \left(-1 + \frac{(-1)^{2}}{2^{2}} + \frac{(-1)^{3}}{3^{2}} + \ldots + \frac{(-1)^{n}}{n^{2}}\right)$. For $n=1$: $a_{1} = \int_{-1}^{1} (1) dx = [x]_{-1}^{1} = 2$. For $n=2$: $a_{2} = \int_{-1}^{2} (1 + \frac{x}{2}) dx = [x + \frac{x^{2}}{4}]_{-1}^{2} = (2 + 1) - (-1 + \frac{1}{4}) = 3 - (-0.75) = 3.75$. For $n=3$: $a_{3} = \int_{-1}^{3} (1 + \frac{x}{2} + \frac{x^{2}}{3}) dx = [x + \frac{x^{2}}{4} + \frac{x^{3}}{9}]_{-1}^{3} = (3 + \frac{9}{4} + 3) - (-1 + \frac{1}{4} - \frac{1}{9}) = 6 + 2.25 - (-0.861) = 9.111$. For $n=4$: $a_{4} = \int_{-1}^{4} (1 + \frac{x}{2} + \frac{x^{2}}{3} + \frac{x^{3}}{4}) dx = [x + \frac{x^{2}}{4} + \frac{x^{3}}{9} + \frac{x^{4}}{16}]_{-1}^{4} = (4 + 4 + \frac{64}{9} + 16) - (-1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16}) \approx 24 + 7.11 + 0.86 = 31.97$. Since $a_{4} > 30$,the values of $n$ for which $a_{n} \in (2, 30)$ are $n=2$ and $n=3$. The sum of these elements is $2 + 3 = 5$.

Let $a_{n} = \int_{-1}^{n} \left(1 + \frac{x}{2} + \frac{x^{2}}{3} + \ldots + \frac{x^{n-1}}{n}\right) dx$ for $n \in N$. Then the sum of all the elements of the set $\{n \in N : a_{n} \in (2, 30)\}$ is $...........$

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