If sum_{r=1}^{n}(2r-1) = x,then find the valu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $x = \sum_{r=1}^{n}(2r-1)$.This is the sum of the first $n$ odd integers,which is $x = n^2$.Therefore,$x^2 = n^4$.The expression become

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Given that $x = \sum_{r=1}^{n}(2r-1)$.This is the sum of the first $n$ odd integers,which is $x = n^2$.Therefore,$x^2 = n^4$.The expression becomes $\lim_{n \rightarrow \infty} \left[ \frac{1^3 + 2^3 + \ldots + n^3}{x^2} \right]$.Using the formula for the sum of cubes,$\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.Substituting this into the limit: $\lim_{n \rightarrow \infty} \left[ \frac{n^2(n+1)^2}{4n^4} \right]$.$= \lim_{n \rightarrow \infty} \left[ \frac{n^2 \cdot n^2(1 + \frac{1}{n})^2}{4n^4} \right]$.$= \lim_{n \rightarrow \infty} \left[ \frac{(1 + \frac{1}{n})^2}{4} \right] = \frac{1}{4}$.

If $\sum_{r=1}^{n}(2r-1) = x$,then find the value of $\lim_{n}$ ${\rightarrow \infty} \left[ \frac{1^3}{x^2} + \frac{2^3}{x^2} + \frac{3^3}{x^2} + \ldots + \frac{n^3}{x^2} \right]$.

Similar Questions

$\lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}} = $

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

If $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$,then

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sum_{k=1}^{n} {k^2}}}{{{n^3}}}} \right] = $

The value of $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 2}}} \right)^{\frac{{x + 1}}{3}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module