The sum to 10 terms of the series frac{1}{1+1 | vedclass

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(B) The general term of the series is $T_r = \frac{r}{1+r^2+r^4}$.We know that $1+r^2+r^4 = (1+r^2)^2 - r^2 = (1+r^2-r)(1+r^2+r)$.Thus,$T_r = \frac{r}

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$\frac{55}{111}$

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(B) The general term of the series is $T_r = \frac{r}{1+r^2+r^4}$.We know that $1+r^2+r^4 = (1+r^2)^2 - r^2 = (1+r^2-r)(1+r^2+r)$.Thus,$T_r = \frac{r}{(r^2-r+1)(r^2+r+1)}$.Using partial fractions,we can write:$T_r = \frac{1}{2} \left[ \frac{1}{r^2-r+1} - \frac{1}{r^2+r+1} \right]$.Let $f(r) = \frac{1}{r^2-r+1}$. Then $T_r = \frac{1}{2} [f(r) - f(r+1)]$.The sum of $10$ terms is $S_{10} = \sum_{r=1}^{10} T_r = \frac{1}{2} [f(1) - f(11)]$.$f(1) = \frac{1}{1^2-1+1} = 1$.$f(11) = \frac{1}{11^2-11+1} = \frac{1}{121-11+1} = \frac{1}{111}$.Therefore,$S_{10} = \frac{1}{2} [1 - \frac{1}{111}] = \frac{1}{2} [\frac{110}{111}] = \frac{55}{111}$.

The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots$ is:

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