Question

Find the general term of the following expressions when expanded in ascending powers of $x$.
$\frac{x^2+7x+3}{x^2+7x+10}$

Answer 1

$\frac{x^2+7x+3}{x^2+7x+10}=\frac{x^2+7x+10-10+3}{x^2+7x+10}=1-\frac{7}{x^2+7x+10}$

Now,

$\frac{7}{x^2+7x+10}=\frac{7}{(x+2)(x+5)}=\frac{A}{x+2}+\frac{B}{x+5}$

$=\frac{Ax+5A+Bx+2B}{(x+2)(x+5)}$

$=\frac{x(A+B)+(5A+2B)}{(x+2)(x+5)}$

Now,

$7=(A+B)x+(5A+2B)$

$\Rightarrow$ $A+B=0$ --- ( 1 )

$\Rightarrow$ $5A+2B=7$ ---- ( 2 )

By solving ( 1 ) and ( 2 ) we get,

$A=\frac{7}{3},B=\frac{-7}{3}$

Now the equation is $=1-\frac{7}{3(x+2)}-\frac{7}{3(x+5)}$

For general term $=-\frac{7}{3(x+2)}-\frac{7}{3(x+5)}=-\frac{7}{3}(x+2{)}^{-1}-\frac{7}{3}(x+5{)}^1$

$=\frac{7\times 1}{3\times 2}{(\frac{x}{2}+1)}^{-1}-\frac{7\times 1}{3\times 5}{(\frac{x}{5}+1)}^{-1}$

$=-\frac{7}{3}\left\{\right(-1{)}^r\frac{1}{2^{r+1}}{x}^r\}-\frac{7}{3}\left\{\right(-1{)}^r\left(\frac{1}{5^{r+1}}\right)x^r\}$

$=-\frac{7}{3}(-1{)}^r\{\frac{1}{5^{r+1}}+\frac{1}{2^{r+1}}\}{x}^r$