Can you evaluate this sum by using the properties of the sigma notation ? Or I must develop this and evaluate them one by one ?
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4You can write $\frac{k^2}{k-1}=\frac{k^2-1+1}{k-1}=k+1+\frac{1}{k-1}$ and then the sum as $S=\sum_{k=6}^{12}{(k+1)} + \sum_{k=6}^{12}{\frac{1}{k-1}}$. The first summation is easy enough, bu the second is a harmonic sum, which has no exact closed-form solution in the general case. You will need to add the terms one by one for the second summation. – Marconius Aug 28 '15 at 23:58
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Ok, so I guess my problem is asking me to do the sum one by one for the initial form also because I thought I would be able to use properties to simplify the work. THank you! – Aug 29 '15 at 00:03