If sum_{k=1}^{10} frac{k}{k^{4}+k^{2}+1}=frac | vedclass

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(A) We have the sum $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}$.Note that $k^{4}+k^{2}+1 = (k^{2}+1)^{2}-k^{2} = (k^{2}+k+1)(k^{2}-k+1)$.Thus,the genera

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$166$

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(A) We have the sum $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}$.Note that $k^{4}+k^{2}+1 = (k^{2}+1)^{2}-k^{2} = (k^{2}+k+1)(k^{2}-k+1)$.Thus,the general term is $\frac{k}{(k^{2}+k+1)(k^{2}-k+1)}$.Using partial fractions,we write $\frac{k}{(k^{2}+k+1)(k^{2}-k+1)} = \frac{1}{2} \left( \frac{1}{k^{2}-k+1} - \frac{1}{k^{2}+k+1} \right)$.Let $f(k) = \frac{1}{k^{2}-k+1}$. Then the sum is $\frac{1}{2} \sum_{k=1}^{10} (f(k) - f(k+1))$.This is a telescoping sum: $\frac{1}{2} (f(1) - f(11))$.$f(1) = \frac{1}{1^{2}-1+1} = 1$.$f(11) = \frac{1}{11^{2}-11+1} = \frac{1}{121-11+1} = \frac{1}{111}$.So,the sum is $\frac{1}{2} (1 - \frac{1}{111}) = \frac{1}{2} (\frac{110}{111}) = \frac{55}{111}$.Here $m=55$ and $n=111$,which are coprime.Therefore,$m+n = 55+111 = 166$.

If $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$,where $m$ and $n$ are coprime,then $m+n$ is equal to.

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