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Question Number 147412 by vvvv last updated on 20/Jul/21

Answered by puissant last updated on 21/Jul/21

$$={lim}_{k\to \mathrm{\infty }}\frac{1}{k}\underset{n=1}{\sum ^k}\frac{n^2+3nk+9k^{2}sin\left(\frac{n}{k}\right)}{k^2}$$$$={lim}_{k\to \mathrm{\infty }}\frac{1}{k}\underset{n=1}{\sum ^k}[{\left(\frac{n}{k}\right)}^2+3\left(\frac{n}{k}\right)+9sin\left(\frac{n}{k}\right)$$$$Riemannintegral..$$$$\Rightarrow {lim}_{k\to \mathrm{\infty }}{U}_k={\int }_0^1{x}^2+3x+9sin\left(x\right)dx$$$$=\frac{1}{3}{\left[x^3\right]}_0^1+\frac{3}{2}{\left[x^2\right]}_0^1-9{\left[cos\left(x\right)\right]}_0^1$$$$=\frac{1}{3}+\frac{3}{2}-9cos\left(1\right)+9$$$$\Rightarrow {lim}_{k\to \mathrm{\infty }}{U}_k=\frac{65}{6}-9cos\left(1\right)..$$

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