lim_(n rarr oo)sum_(i=1)^(n)(16 i)/(n^(2))

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Recognize Riemann Sum: We recognize that the given series is a Riemann sum. The Riemann sum is used to approximate the integral of a function over an interval by partitioning the interval and summing up areas of rectangles under the curve. The general form of a Riemann sum is sum_(i=1)^(n)f(x_i)*(Delta x), where Delta x is the width of each partition and f(x_i) is the function value at the ith partition. In our case, f(x_i) = 16x, and Delta x = 1/n. We can rewrite the series as sum_(i=1)^(n)(16i/n^2) = sum_(i=1)^(n)(16/n^2)*i. Here, 16/n^2 is constant for each term in the sum, and i represents the ith partition's contribution to the sum.Express as Riemann Sum: We can now express the sum as a Riemann sum that approximates the integral of the function f(x) = 16x from 0 to 1 as n approaches infinity. The Riemann sum becomes lim_(n rarr oo)(16/n^2) * sum_(i=1)^(n)i. The sum sum_(i=1)^(n)i is the sum of the first n natural numbers, which can be calculated using the formula n(n+1)/2.Substitute Formula: Substitute the formula for the sum of the first n natural numbers into the Riemann sum: lim_(n rarr oo)(16/n^2) * (n(n+1)/2). This simplifies to lim_(n rarr oo)(16/n^2) * (n^2/2 + n/2).Distribute Terms: Distribute 16/n^2 across the terms in the parentheses: lim_(n rarr oo)(16/2 + 16n/(2n^2)). This simplifies to lim_(n rarr oo)(8 + 8/n).Take Limit: Now, take the limit as n approaches infinity. The term 8/n approaches 0 as n becomes very large, so the limit is lim_(n rarr oo)(8 + 8/n) = 8 + 0.Final Answer: The final answer is the limit of the sum, which is 8. This is the value of the integral of f(x) = 16x from 0 to 1.

$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}$

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$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}$