Question

Let $r$ be the positive and zero of $P(x)=9x^5+$ $7x^2-9$. If the sum $S=r^4+2r^9+3r^{14}+4r^{19}+$ $\dots \infty$ can be expressed as the rational number $\left(\frac{a}{b}\right)$ in the lowest term, then find the sum of digits in $(a+b)$

Answer 1

Answer: 0004

$P(r)=9r^5+7r^2-9=0$
$\therefore 9\left(1-r^5\right)=7r^2...(i)$
Also, $S=r^4+2r^9+3r^{14}+4r^{19}+\dots \infty$
$\frac{S\cdot r^5=+r^9+2r^{14}+3r^{19}+\dots ..\infty }{S\left(1-r^5\right)=r^4+r^9+r^{14}+\dots \infty }$
$=\frac{r^4}{1-r^5}$
$S=\frac{r^4}{{\left(1-r^5\right)}^2}$
Using $1-r^5=\frac{7r^2}{9}$
$S=\frac{81}{49}=\frac{a}{b}$
$\Rightarrow (a+b)=130$
$\therefore$ Sum of digits $=4$