Mathematics · 2022
JEE Main · 24 June 2022, Shift 2 · Q7
lim_n → ∞(n^2/((n^2+1)(n+1))+n^2/((n^2+4)(n+2))+n^2/((n^2+9)(n+3))+…+n^2/((n^2+n^2)(n+n))) is equal to
$$\lim _{n \rightarrow \infty}\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\frac{n^2}{\left(n^2+9\right)(n+3)}+\ldots+\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)
$$
is equal to
Official answer
(1)
$\displaystyle \frac{\pi}{8}+\frac{1}{4} \log _e 2$
More from Definite Integration
- If the value of the integral ∫_0^5 (x+[x])/(e^x-[x]) d x=α e^-1+β, where α, β ∈ R, 5 α+6 β=0, and [x] denotes the greatest integer less than or equal…2021
- If for all real triplets (a, b, c), f(x)=a+b x+cx^2; then ∫_0^1 f(x) d x is equal to:2020
- If the value of the integral ∫_0^1 / 2 x^2/((1-x^2)^3 / 2) d x is k/6, then k is equal to:2020
- Let 0 ≤ x ≤ 2Max{(9-x^2)/(5-x)}=α and 0 ≤ x ≤ 2Min{(9-x^2)/(5-x)}=β. If ∫_β-8/3^2 α-1 Max{(9-x^2)/(5-x), x} d x=α_1+α_2 log_e(8/15) then α_1+α_2 is…2022
- Let the function f:[0,2] → R be defined as f(x)= {e^min {x^2, x-[x]}, x ∈[0,1); e^[x- log_e x], x ∈[1,2]} where [t] denotes the greatest integer less…2023
- If I_m, n=∫_0^1 x^m-1(1-x)^n-1 d x, for m, n ≥ 1, and ∫_0^1 (x^m-1+x^n-1)/((1+x)^m+n) d x=α I_m, n, α ∈ R, then α equals \ \ \ \.2021
- If [·] represents the greatest integer function, then the value of |∫_0^√((π)/2)[[x^2]- cos x] d x| is \ \ \ \.2021
- The value of k ∈ N for which the integral I_n=∫_0^1(1-x^k)^n d x, n ∈ N, satisfies 147 I_20=148 I_21 is2024
JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.