JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 21)

Let $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3} + \,\,.....\,\, + \,\,{{{x^{n - 1}}} \over n}} \right)dx} $$ for every n $$\in$$ N. Then the sum of all the elements of the set {n $$\in$$ N : an $$\in$$ (2, 30)} is ____________.
Answer
5

Explanation

$$\because$$ $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3}\, + \,....\, + \,{{{x^{n - 1}}} \over n}} \right)dx} $$

$$ = \left[ {x + {{{x^2}} \over {{2^2}}} + {{{x^3}} \over {{3^2}}}\, + \,......\, + \,{{{x^n}} \over {{n^2}}}} \right]_{ - 1}^n$$

$${a_n} = {{n + 1} \over {{1^2}}} + {{{n^2} - 1} \over {{2^2}}} + {{{n^3} + 1} \over {{3^2}}} + {{{n^4} - 1} \over {{4^2}}}\, + \,...\, + \,{{{n^n} + {{( - 1)}^{n + 1}}} \over {{n^2}}}$$

Here, $${a_1} = 2,\,{a_2} = {{2 + 1} \over 1} + {{{2^2} - 1} \over 2} = 3 + {3 \over 2} = {9 \over 2}$$

$${a_3} = 4 + 2 + {{28} \over 9} = {{100} \over 9}$$

$${a_4} = 5 + {{15} \over 4} + {{65} \over 9} + {{255} \over {16}} > 31$$.

$$\therefore$$ The required set is $$\{ 2,3\} $$. $$\because$$ $${a_n} \in (2,30)$$

$$\therefore$$ Sum of elements = 5.

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