If the sum of the first ten terms of the seri | vedclass

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(C) The general term of the series is $T_{n} = \frac{n}{4n^{4}+1}$.We can rewrite the denominator as: $4n^{4}+1 = (2n^{2}+1)^{2} - (2n)^{2} = (2n^{2}+

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$276$

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(C) The general term of the series is $T_{n} = \frac{n}{4n^{4}+1}$.We can rewrite the denominator as: $4n^{4}+1 = (2n^{2}+1)^{2} - (2n)^{2} = (2n^{2}+2n+1)(2n^{2}-2n+1)$.Using partial fractions:$T_{n} = \frac{1}{4} \left[ \frac{1}{2n^{2}-2n+1} - \frac{1}{2n^{2}+2n+1} ight]$.Let $f(n) = \frac{1}{2n^{2}-2n+1}$. Then $T_{n} = \frac{1}{4} [f(n) - f(n+1)]$.The sum of the first $10$ terms is:$S_{10} = \sum_{n=1}^{10} T_{n} = \frac{1}{4} [f(1) - f(11)]$.$f(1) = \frac{1}{2(1)^{2}-2(1)+1} = 1$.$f(11) = \frac{1}{2(11)^{2}-2(11)+1} = \frac{1}{242-22+1} = \frac{1}{221}$.$S_{10} = \frac{1}{4} [1 - \frac{1}{221}] = \frac{1}{4} 	imes \frac{220}{221} = \frac{55}{221}$.Since $m = 55$ and $n = 221$ are co-prime,$m + n = 55 + 221 = 276$.

If the sum of the first ten terms of the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\ldots$ is $\frac{m}{n}$,where $m$ and $n$ are co-prime numbers,then $m + n$ is equal to

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