If the sum of the first 20 terms of the serie | vedclass

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(C) The $r$-th term of the series is $T_r = \frac{4r}{4+3r^2+r^4}$.We can factorize the denominator: $r^4+3r^2+4 = (r^4+4r^2+4)-r^2 = (r^2+2)^2 - r^2

Vedclass · Accepted Answer

$421$

Vedclass · Answer

(C) The $r$-th term of the series is $T_r = \frac{4r}{4+3r^2+r^4}$.We can factorize the denominator: $r^4+3r^2+4 = (r^4+4r^2+4)-r^2 = (r^2+2)^2 - r^2 = (r^2+r+2)(r^2-r+2)$.Thus,$T_r = \frac{4r}{(r^2+r+2)(r^2-r+2)}$.Using partial fractions: $T_r = 2 \left( \frac{1}{r^2-r+2} - \frac{1}{r^2+r+2} \right)$.Let $f(r) = \frac{1}{r^2-r+2}$. Then $f(r+1) = \frac{1}{(r+1)^2-(r+1)+2} = \frac{1}{r^2+2r+1-r-1+2} = \frac{1}{r^2+r+2}$.So,$T_r = 2(f(r) - f(r+1))$.The sum $S_{20} = \sum_{r=1}^{20} 2(f(r) - f(r+1)) = 2(f(1) - f(21))$.$f(1) = \frac{1}{1^2-1+2} = \frac{1}{2}$.$f(21) = \frac{1}{21^2-21+2} = \frac{1}{441-21+2} = \frac{1}{422}$.$S_{20} = 2 \left( \frac{1}{2} - \frac{1}{422} \right) = 1 - \frac{2}{422} = \frac{420}{422} = \frac{210}{211}$.Here $m = 210$ and $n = 211$,which are coprime.Thus,$m+n = 210+211 = 421$.

If the sum of the first $20$ terms of the series $\frac{4.1}{4+3.1^2+1^4}+\frac{4.2}{4+3.2^2+2^4}+\frac{4.3}{4+3.3^2+3^4}+\frac{4.4}{4+3.4^2+4^4}+\ldots$ is $\frac{m}{n}$,where $m$ and $n$ are coprime,then $m + n$ is equal to :-

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