Question ID: #1035
If the sum of the first $10$ terms of the series $\frac{4 \cdot 1}{1+4 \cdot 1^{4}}+\frac{4 \cdot 2}{1+4 \cdot 2^{4}}+\frac{4 \cdot 3}{1+4 \cdot 3^{4}}+\dots$ is $\frac{m}{n}$ where $\gcd(m,n)=1$, then $m+n$ is equal to
Solution:
General term of the given series is:
$$T_r = \frac{4r}{1+4r^4}$$
Factorize the denominator using the identity $a^2-b^2 = (a-b)(a+b)$:
$$1+4r^4 = (2r^2+1)^2 – 4r^2$$
$$1+4r^4 = (2r^2 – 2r + 1)(2r^2 + 2r + 1)$$
Substitute the factors back into the general term:
$$T_r = \frac{4r}{(2r^2 – 2r + 1)(2r^2 + 2r + 1)}$$
Express the numerator as the difference of the two factors in the denominator:
$$T_r = \frac{(2r^2+2r+1) – (2r^2-2r+1)}{(2r^2-2r+1)(2r^2+2r+1)}$$
$$T_r = \frac{1}{2r^2 – 2r + 1} – \frac{1}{2r^2 + 2r + 1}$$
Substitute $r = 1, 2, 3, \dots, 10$ to find the individual terms:
$$T_1 = \frac{1}{2(1)^2 – 2(1) + 1} – \frac{1}{2(1)^2 + 2(1) + 1} = \frac{1}{1} – \frac{1}{5}$$
$$T_2 = \frac{1}{2(2)^2 – 2(2) + 1} – \frac{1}{2(2)^2 + 2(2) + 1} = \frac{1}{5} – \frac{1}{13}$$
$$T_3 = \frac{1}{2(3)^2 – 2(3) + 1} – \frac{1}{2(3)^2 + 2(3) + 1} = \frac{1}{13} – \frac{1}{25}$$
$$\dots$$
$$T_{10} = \frac{1}{2(10)^2 – 2(10) + 1} – \frac{1}{2(10)^2 + 2(10) + 1} = \frac{1}{181} – \frac{1}{221}$$
Summing all the terms, the intermediate terms cancel out (telescoping series):
$$S_{10} = T_1 + T_2 + T_3 + \dots + T_{10}$$
$$S_{10} = \left(\frac{1}{1} – \frac{1}{5}\right) + \left(\frac{1}{5} – \frac{1}{13}\right) + \dots + \left(\frac{1}{181} – \frac{1}{221}\right)$$
$$S_{10} = 1 – \frac{1}{221} = \frac{220}{221}$$
Given that $S_{10} = \frac{m}{n}$ and $\gcd(m,n) = 1$:
$$m = 220$$
$$n = 221$$
Calculate the required value:
$$m + n = 220 + 221 = 441$$
Ans. (441)
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