Let \(a_{n}=\int_{-1}^{n}\left(1+\frac{x}{2}+\frac{x^{2}}{3}+\frac{x^{3}}{4}+\ldots+\frac{x^{n-1}}{n}\right) d x\) for \(n \in N\). Then the sum of all the elements of the set \(\{n \in N: a_{n} \in(2,30)\}\) is

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Definite Integrals

Let

a_{n} = \int_{- 1}^{n} (1 + \frac{x}{2} + \frac{x ^{2}}{3} + \frac{x ^{3}}{4} + \dots + \frac{x ^{n - 1}}{n}) d x

for

n \in N

. Then the sum of all the elements of the set

{n \in N : a_{n} \in (2, 30)}

Answer: 5

Writing the integrand as the finite sum

\sum_{k = 0}^{n - 1} \frac{x ^{k}}{k + 1}

and integrating term by term gives

a_{n} = \sum_{m = 1}^{n} \frac{n ^{m} - ( - 1 ) ^{m}}{m ^{2}}

. Evaluating the first few values,

a_{1} = \int_{- 1}^{1} 1 d x = 2

, which fails to qualify because the interval

(2, 30)

is open;

a_{2} = \int_{- 1}^{2} (1 + \frac{x}{2}) d x = \frac{15}{4} = 3.75

lies inside it;

a_{3} = \frac{82}{9} \approx 9.11

also lies inside; while

a_{4} = \frac{4595}{144} \approx 31.9

already exceeds

30

, and the terms only grow for larger

n

. Thus the qualifying set is

{2, 3}

and the sum of its elements is

5

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer