The sum of the series frac{1}{1-3 cdot 1^2+1^ | vedclass

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(D) The general term of the series is $T_r = \frac{r}{1-3r^2+r^4}$.We can rewrite the denominator as $r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2-1

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$-\frac{55}{109}$

Vedclass · Answer

(D) The general term of the series is $T_r = \frac{r}{1-3r^2+r^4}$.We can rewrite the denominator as $r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2-1)^2 - r^2$.Using the difference of squares formula,$a^2 - b^2 = (a-b)(a+b)$,we get $(r^2-r-1)(r^2+r-1)$.Thus,$T_r = \frac{r}{(r^2-r-1)(r^2+r-1)}$.Using partial fractions,we 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 $\sum_{r=1}^{10} T_r = \frac{1}{2} [f(1) - f(11)]$.$f(1) = \frac{1}{1^2-1-1} = -1$.$f(11) = \frac{1}{11^2+11-1} = \frac{1}{121+11-1} = \frac{1}{131}$ is incorrect; evaluating $f(11)$ correctly: $f(11) = \frac{1}{11^2+11-1} = \frac{1}{121+11-1} = \frac{1}{131}$ is wrong. Let's re-evaluate: $f(r+1) = \frac{1}{(r+1)^2+(r+1)-1} = \frac{1}{r^2+2r+1+r+1-1} = \frac{1}{r^2+3r+1}$.Actually,$f(r+1) = \frac{1}{(r+1)^2-(r+1)-1} = \frac{1}{r^2+2r+1-r-1-1} = \frac{1}{r^2+r-1}$.So,$\sum_{r=1}^{10} T_r = \frac{1}{2} [f(1) - f(11)] = \frac{1}{2} [-1 - \frac{1}{11^2+11-1}] = \frac{1}{2} [-1 - \frac{1}{121+11-1}] = \frac{1}{2} [-1 - \frac{1}{131}] = \frac{1}{2} [-\frac{132}{131}] = -\frac{66}{131}$.Wait,checking the original series: $r=10 \implies f(11) = \frac{1}{10^2+10-1} = \frac{1}{109}$.Sum $= \frac{1}{2} [-1 - \frac{1}{109}] = \frac{1}{2} [-\frac{110}{109}] = -\frac{55}{109}$.

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

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