If the sum of the first 10 terms of the series $\frac{1}{1 + 1^{4} imes 4} + \frac{2}{1 + 2^{4}

Question

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

Accepted Answer

Sol. [Key Concept: Factor $1 + 4r^{4} = (2r^{2} + 2r + 1)(2r^{2} - 2r + 1)$; the term telescopes, so sum the first ten and read off $m + n$.]$${T_{r} = \frac{r}{1 + 4r^{4}} = \frac{r}{4r^{4} + 4r^{2} + 1 - 4r^{2}} = \frac{r}{\left( 2r^{2} + 1 ight)^{2} - (2r)^{2}} }{= \frac{r}{\left( 2r^{2} + 2r + 1 ight)\left( 2r^{2} - 2r + 1 ight)}}$$$$ {S_{10} = \frac{1}{4}\left( \frac{1}{1} - \frac{1}{221} ight) = \frac{55}{221} = \frac{m}{n}}$$$$ {	herefore m + n = 276 }$$

Question

Options

Solution

More Progression (Sequence and Series) Questions