Question

Let $r$ be the positive and zero of $P(x)=9 x^{5}+$ $7 x ^{2}-9$. If the sum $S = r ^{4}+2 r ^{9}+3 r ^{14}+4 r ^{19}+$ $\ldots \infty$ 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

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