MCQ
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to
  • A
    $12$
  • B
    $\frac{19}{3}$
  • C
    $0$
  • $19$

Answer

Correct option: D.
$19$
d
$\lim _{n \rightarrow \infty} \frac{3}{n} \sum \limits_{r=0}^{n-1}\left(2+\frac{r}{n}\right)^2$

$=3 \int \limits_0^1(2+x)^2 d x=27-8=19$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:
  1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
  2. $\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
  3. $\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
  4. $\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$ 
Points within and on the boundary of the feasible region represents feasible solutions of the constraints. Any point outside the feasible region is an
The shortest distance between the lines $\frac{{x - 3}}{2} = \frac{{y + 15}}{{ - 7}} = \frac{{z - 9}}{5}$ and $\frac{{x + 1}}{2} = \frac{{y - 1}}{1} = \frac{{z - 9}}{{ - 3}}$ is
Let $\vec a = 2\hat i - \hat j + \hat k$, $\vec b = \hat i + 2\hat j - \hat k$ and $\vec c = \hat i + \hat j - 2\hat k$ be three vectors. A vector of the type $\vec b + \lambda \vec c$ for some scalar $\lambda $,  whose projection on $\vec a$ is of magnitude $\sqrt {\frac{2}{3}} $ is
The vector component of $\vec{\text{b}}$ perpendicular to $\vec{\text{a}}$ is:
  1. $\big(\vec{\text{b}}.\vec{\text{c}}\big)\vec{\text{a}}$
  2. $\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$
  3. $\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
  4. None of these
If $A, B, C$  be the angles of a triangle, then $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}\,} \right| = $
The value of $b$ and $c$ for which the identity $f(x + 1) - f(x) = 8x + 3$ is satisfied, where $f(x) = b{x^2} + cx + d$, are
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then f(x) is continuous at $\text{x}=\frac{\pi}{2},$ if:
  1. $\text{a}=\frac{1}{3},\text{ b}=2$
  2. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
  3. $\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
  4. none of these
If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt x )^2$, then
The solution of the differential equation $\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is