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

Question

If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1+4 \cdot 1^4}+\frac{4 \cdot 2}{1+4 \cdot 2^4}+\frac{4 \cdot 3}{1+4 \cdot 3^4}+\ldots .$. is $\frac{\mathrm{m}}{\mathrm{n}}$

Accepted Answer

$\begin{gathered} \frac{4.1}{1+{4.1}^{4}}+\frac{4.2}{1+{4.2}^{4}}+\frac{4.3}{1+{4.3}^{4}}+\ldots . \ {T}_{r}=\frac{4r}{1+4{r}^{4}}=\frac{4r}{4{r}^{4}+4{r}^{2}+1-4{r}^{2}} \ =\frac{4r}{{(2{r}^{2}+1)}^{2}-(2r{)}^{2}} \ {T}_{r}=\frac{4r}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)} \end{gathered}$

$\begin{gathered} {T}_{r}=\frac{(2{r}^{2}+2r+1)-(2{r}^{2}-2r+1)}{(2{r}^{2}-2r+1)(2{r}^{2}+2r+1)} \ {T}_{r}=(\frac{1}{{r}^{2}+(r-1{)}^{2}}-\frac{1}{{r}^{2}+(r+1{)}^{2}}) \ \sum _{r=1}^{10}{T}_{r}=(\frac{1}{{0}^{2}+{1}^{2}}-\frac{1}{{1}^{2}+{2}^{2}}+\frac{1}{{1}^{2}+{2}^{2}}-\frac{1}{{2}^{2}+{3}^{2}}+\ldots . \ \frac{1}{{9}^{2}+{10}^{2}}-\frac{1}{{10}^{2}+{11}^{2}} \ =1-\frac{1}{221} \ =\frac{220}{221} \ 	herefore \, m+n=220+221 \ \, =441 \end{gathered}$

Solutions · 1