If the sum of the first 10 terms of the serie | vedclass

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(B) The $r$-th term of the series is given by $T_r = \frac{4r}{1+4r^4}$.Using the identity $1+4r^4 = (1+2r^2-2r)(1+2r^2+2r)$,we can write:$T_r = \frac

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$441$

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(B) The $r$-th term of the series is given by $T_r = \frac{4r}{1+4r^4}$.Using the identity $1+4r^4 = (1+2r^2-2r)(1+2r^2+2r)$,we can write:$T_r = \frac{(2r^2+2r+1)-(2r^2-2r+1)}{(2r^2+2r+1)(2r^2-2r+1)}$$T_r = \frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1}$.For $r=1, T_1 = \frac{1}{1} - \frac{1}{5}$.For $r=2, T_2 = \frac{1}{5} - \frac{1}{13}$.Continuing this up to $r=10$,$T_{10} = \frac{1}{2(10)^2-2(10)+1} - \frac{1}{2(10)^2+2(10)+1} = \frac{1}{181} - \frac{1}{221}$.The sum $S_{10} = \sum_{r=1}^{10} T_r = 1 - \frac{1}{221} = \frac{220}{221}$.Given $S_{10} = \frac{m}{n}$,we have $m=220$ and $n=221$.Since $\operatorname{gcd}(220, 221) = 1$,$m+n = 220+221 = 441$.

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .

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