If the sum of the first 10 terms of the series [(1)/(1+1^4) + (2)/(1+2^4) + (3)/(1+3^4) + …

Question

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

Accepted Answer

The general term is $T_r = \dfrac{r}{1+4r^4}$. Since $1+4r^4 = (2r^2-2r+1)(2r^2+2r+1)$, and with $a_r = 2r^2-2r+1$ we get $a_{r+1}=2r^2+2r+1$, so
$$T_r = \frac14\left(\frac{1}{a_r}-\frac{1}{a_{r+1}}\right).$$

Telescoping:
$$S_{10} = \frac14\left(\frac{1}{a_1}-\frac{1}{a_{11}}\right) = \frac14\left(1-\frac{1}{221}\right) = \frac14\cdot\frac{220}{221} = \frac{55}{221}.$$

Since $\gcd(55,221)=1$, $m=55$, $n=221$, so $m+n = 276$.

Solutions · 1