Mathematics, 27.08.2020 23:01 Pattondestiny
Use the summation formulas to rewrite the expression without the summation notation. summation_k = 1^n 12k(k - 1) / n^3 Use the result to find the sums for n = 10, n= 100, n= 1000 n= 10,000.
Answers: 1
![Sum=4*[\frac{n^2-1}{n^2}]](/tpl/images/0731/9121/ad551.png)
Eq (1)
![=\frac{12}{n^3}[\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}]\\Taking\ n\ as\ common\\=n*\frac{12}{n^3}[\frac{(n+1)(2n+1)}{6}-\frac{(n+1)}{2}] \\=\frac{12}{n^2}*[\frac{(n+1)(2n+1)}{6}]-\frac{12}{n^2}*[\frac{(n+1)}{2}] \\=\frac{2*(n+1)(2n+1)}{n^2}-\frac{6(n+1)}{n^2}\\](/tpl/images/0731/9121/dc2f6.png)
as common: 
![Sum=4[\frac{(10^{2})-1}{10^{2}}]\\Sum=3.96](/tpl/images/0731/9121/e7a34.png)
![Sum=4[\frac{(100^{2})-1}{100^{2}}]\\Sum=3.9996](/tpl/images/0731/9121/c2426.png)
![Sum=4[\frac{(1000^{2})-1}{1000^{2}}]\\Sum=3.999996](/tpl/images/0731/9121/5ce13.png)
![Sum=4[\frac{(10000^{2})-1}{10000^{2}}]\\Sum=3.99999996](/tpl/images/0731/9121/ff9dc.png)
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Use the summation formulas to rewrite the expression without the summation notation. summation_k = 1...
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