The value of limn→∞ $$1n4($$∑$$r=12n(3nr2+2n2r))$$ is equal to

limn→∞ $$1n4$$∑$$r=12n3nr2+2n2r$$$$=limn$$→∞ $$1n4n3$$ ∑$$r=12n3rn2+2rn$$ $$=$$∫$$023x2+2xdx$$ $$=x3+x202$$ $$=8+4=12$$

Evaluation of definite integrals

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Question

$The value of \underset{n \to \infty}{l i m} \frac{1}{n^{4}} (\sum_{r = 1}^{2 n} (3 n r^{2} + 2 n^{2} r)) i s e q u a l t o$

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Solution

$\begin{array}{l} \underset{n \to \infty}{l i m} \frac{1}{n^{4}} \sum_{r = 1}^{2 n} (3 n r^{2} + 2 n^{2} r) = \underset{n \to \infty}{l i m} \frac{1}{n^{4}} n^{3} \sum_{r = 1}^{2 n} (3 {(\frac{r}{n})}^{2} + 2 (\frac{r}{n})) \\ = \int_{0}^{2} (3 x^{2} + 2 x) d x \\ = {(x^{3} + x^{2})}_{0}^{2} \\ = 8 + 4 = 12 \end{array}$

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Evaluation of definite integrals

Remember concepts with our Masterclasses.

The value of limn→∞ 1n4(∑r=12n(3nr2+2n2r)) is equal to

limn→∞ 1n4∑r=12n3nr2+2n2r​=limn→∞ 1n4n3 ∑r=12n3rn2+2rn =∫023x2+2xdx ​=x3+x202​ =8+4=12

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$The value of \underset{n \to \infty}{l i m} \frac{1}{n^{4}} (\sum_{r = 1}^{2 n} (3 n r^{2} + 2 n^{2} r)) i s e q u a l t o$