Solveeit Logo
Login

Question

Question: Let a =![](https://cdn.pureessence.tech/canvas_151.png?top_left_x=277&top_left_y=1662&width=300&heig...

Let a =∑r=1nr2n3\frac{\sum_{r = 1}^{n}r^{2}}{n^{3}} and b = ∑r=1n(r3−r2)n4\frac{\sum_{r = 1}^{n}{(r^{3} - r^{2})}}{n^{4}} then

A

a = b

B

a<b

C

4a – 3b = 0

D

3a – 4b = 0

Answer

3a – 4b = 0

Explanation

Solution

a = lim⁡n→∞\lim _ { n \rightarrow \infty } ∑r=1nr2n3\frac{\sum_{r = 1}^{n}r^{2}}{n^{3}}

a = lim⁡n→∞\lim_{n \rightarrow \infty} ∑r=1n(rn)2×1n\sum_{r = 1}^{n}{\left( \frac{r}{n} \right)^{2} \times \frac{1}{n}}

a = ∫01x2dx\int_{0}^{1}{x^{2}dx}= [x33]01\left\lbrack \frac{x^{3}}{3} \right\rbrack_{0}^{1}

a = 13\frac{1}{3}

b = ∑r=1nr3n4−∑r=1nr2n4\sum_{r = 1}^{n}\frac{r^{3}}{n^{4}} - \sum_{r = 1}^{n}\frac{r^{2}}{n^{4}} ® 0

[Sum of r2 given the expression in n3]

b = ∑r=1nr3n4\sum_{r = 1}^{n}\frac{r^{3}}{n^{4}} = ∑r=1n(rn)3×1n\sum_{r = 1}^{n}{\left( \frac{r}{n} \right)^{3} \times \frac{1}{n}}

b = ∫01x3\int_{0}^{1}x^{3}dx = [x44]01\left\lbrack \frac{x^{4}}{4} \right\rbrack_{0}^{1} = ¼