Let r be the positive real zero of P(x)=9 x5+7 x2-9. If the sum S=r4+2 r9+3 r14+4 r19+..... ∞ can be expressed as the rational number (( a / b )) in the lowest term, then find the sum of digits in ( a + b ).

Question

Let r be the positive real zero of P(x)=9 x5+7 x2-9. If the sum S=r4+2 r9+3 r14+4 r19+..... ∞ can be expressed as the rational number (( a / b )) in the lowest term, then find the sum of digits in ( a + b ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/55a0235983ed30250c1ef6ac38f7b9e1-.png" /> =(r4/1-r5) S=(r4/(1-r5)2) Using 1-r5=(7 r2/9) S =(81/49)=( a / b ) ⇒( a + b )=130 ∴ text Sum of digits =4

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

$= \frac{r ^{4}}{1 - r ^{5}}$
$S = \frac{r ^{4}}{( 1 - r ^{5} ) ^{2}}$
Using $1 - r^{5} = \frac{7 r ^{2}}{9}$
$S = \frac{81}{49} = \frac{a}{b} \Rightarrow (a + b) = 130$
$∴ Sum of digits = 4$

Q. Let r be the positive real zero of P(x)=9x5+7x2−9. If the sum S=r4+2r9+3r14+4r19+..... ∞ can be expressed as the rational number (ba​) in the lowest term, then find the sum of digits in (a+b).

=1−r5r4​ S=(1−r5)2r4​ Using 1−r5=97r2​ S=4981​=ba​⇒(a+b)=130 ∴ Sum of digits =4

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

$= \frac{r ^{4}}{1 - r ^{5}}$
$S = \frac{r ^{4}}{( 1 - r ^{5} ) ^{2}}$
Using $1 - r^{5} = \frac{7 r ^{2}}{9}$
$S = \frac{81}{49} = \frac{a}{b} \Rightarrow (a + b) = 130$
$∴ Sum of digits = 4$