How do you use the summation formulas to rewrite the expression #Sigma (6k(k-1))/n^3# as k=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?
Let
# S_n = sum_(k=1)^n 1/n^3 6k(k-1) #
# :. S_n = 6/n^3sum_(k=1)^n (k^2-k) #
# :. S_n = 6/n^3{sum_(k=1)^n (k^2) - sum_(j=1)^n (k)} #
And using the standard results: We have; And this has been calculated using Excel for
What happens as [ NB As an additional task we could possibly conclude that as Now, And so,
Which confirms our assumption!
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You can rewrite the expression as ( \frac{6}{n^3} \sum_{k=1}^{n} k(k-1) ). The sum ( \sum_{k=1}^{n} k(k-1) ) can be simplified to ( \frac{n(n+1)(2n+1)}{6} - n ). Substituting this into the original expression gives ( \frac{6}{n^3} \left( \frac{n(n+1)(2n+1)}{6} - n \right) ). This simplifies to ( \frac{(n+1)(2n+1)}{n^2} - \frac{6}{n^2} ). For n=10, 100, 1000, and 10000, the sums are approximately 0.67, 0.6734, 0.673467, and 0.67346733, respectively.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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