Let $a _{ n }=\int\limits_{-1}^{ n }\left(1+\frac{ x }{2}+\frac{ x ^2}{2}+\frac{ x ^3}{3}+\ldots \ldots +\frac{ x ^{ n -1}}{ n }\right) dx$ for $n \in N$ Then the sum of all the elements of the set $\left\{ n \in N : a _{ n } \in(2,30)\right\}$ is ______
Correct Answer: 5
Solution and Explanation
The correct answer is 5.
\(\int_{-1}^{n}(1+\frac{x}{2}+\frac{x^{2}}{3}+....+\frac{x^{n-1}}{n})dx\)
\([x+\frac{x^{2}}{2}+\frac{x^{3}}{3^{2}}+...\frac{x^{n}}{n^{2}}]^{n}\)
\((n+\frac{n^{2}}{2^{2}}+\frac{n^{3}}{3^{2}}+...+-\frac{n^{n}}{n^{2}})\)
\(-(-1+\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{4^{2}}+...+\frac{(-1)^{n}}{n^{2}})\)
\(a_{n}=(n+1)+\frac{1}{2^{2}}(n^{2}-1)+\frac{1}{3^{2}}(n^{3}+1)+...+\frac{1}{n^{2}}(n^{n}-(-1)^{n})\)
\(If \; n=1\Rightarrow a_{n}=2\not{\epsilon }(2,30)\)
If n = 2
\(\Rightarrow a_{n}=(2+1)+\frac{1}{2^{2}}(2^{2}-1)=3+\frac{3}{4}< 30\)
If n = 3
\(\Rightarrow a_{n}=(3+1)+\frac{1}{4}(8)+\frac{1}{9}(28)=11+\frac{28}{9}< 30\)
If n = 4
\(\Rightarrow a_{n}=(4+1)+\frac{1}{4}(16-1)+\frac{1}{9}(64+1)+\frac{1}{16}\)
\(=5+\frac{15}{4}+\frac{65}{9}+\frac{255}{16}> 30\)
Test {2,3} sum of elements 5
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).
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