If the sum of the first 10 terms of the series [(41)/(1+41^4) + (42)/(1+42^4) + (43)/(1+43…

Question

If the sum of the first 10 terms of the series
$$\frac{4\cdot1}{1+4\cdot1^4} + \frac{4\cdot2}{1+4\cdot2^4} + \frac{4\cdot3}{1+4\cdot3^4} + \cdots$$
is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 441

Accepted Answer

$T_r = \dfrac{4r}{1+4r^4} = \dfrac{4r}{(2r^2-2r+1)(2r^2+2r+1)} = \dfrac{1}{(r-1)^2+r^2} - \dfrac{1}{r^2+(r+1)^2}$, a telescoping form.

$$\sum_{r=1}^{10}T_r = 1 - \frac{1}{221} = \frac{220}{221}.$$

Since $\gcd(220,221)=1$, $m+n = 220+221 = 441$.

Solutions · 1