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

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(C) The general term $T_k = \frac{k}{1+4k^4}$.We can rewrite the denominator as $1+4k^4 = 1+4k^4+4k^2-4k^2 = (2k^2+1)^2 - (2k)^2 = (2k^2-2k+1)(2k^2+2k

Vedclass · Accepted Answer

$276$

Vedclass · Answer

(C) The general term $T_k = \frac{k}{1+4k^4}$.We can rewrite the denominator as $1+4k^4 = 1+4k^4+4k^2-4k^2 = (2k^2+1)^2 - (2k)^2 = (2k^2-2k+1)(2k^2+2k+1)$.Using partial fractions, $T_k = \frac{1}{4} \left[ \frac{1}{2k^2-2k+1} - \frac{1}{2k^2+2k+1} ight]$.Let $f(k) = 2k^2-2k+1$. Then $f(k+1) = 2(k+1)^2-2(k+1)+1 = 2k^2+4k+2-2k-2+1 = 2k^2+2k+1$.Thus, $T_k = \frac{1}{4} [\frac{1}{f(k)} - \frac{1}{f(k+1)}]$.The sum of the first $10$ terms is $S_{10} = \sum_{k=1}^{10} T_k = \frac{1}{4} \sum_{k=1}^{10} [\frac{1}{f(k)} - \frac{1}{f(k+1)}]$.This is a telescoping sum: $S_{10} = \frac{1}{4} [\frac{1}{f(1)} - \frac{1}{f(11)}]$.$f(1) = 2(1)^2-2(1)+1 = 1$.$f(11) = 2(11)^2-2(11)+1 = 242-22+1 = 221$.$S_{10} = \frac{1}{4} [1 - \frac{1}{221}] = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$.Since $	ext{gcd}(55, 221) = 1$, we have $m=55$ and $n=221$.Therefore, $m+n = 55+221 = 276$.

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

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