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

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(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 $\lim_{n}$ ${\rightarrow \infty} \left[ \frac{1^3}{x^2} + \frac{2^3}{x^2} + \frac{3^3}{x^2} + \ldots + \frac{n^3}{x^2} \right]$.

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