Q. The value of the sum $\sum_{r = 1}^{n}{\sum_{s = 1}^{n}{S_{rs}{\ 2}^{r}3^{s},}whereS_{rs} = 0,\ if\ r \neq s\ andS_{rs} = 1,\ \ if\ r = s,}$ is
A) $\frac{\left( 5^{n} - 1 \right)}{4}$
B) $\frac{6}{5}\left( 6^{n} - 1 \right)$
C) $\frac{5^{n}6^{n}}{n + 1}$
D) $\frac{5}{4}\left( 5^{n} - 1 \right)$
Explanation
(b)
$\sum_{r = 1}^{n}{\sum_{s = 1}^{n}{S_{rs}2^{r}3^{s} = 2 \bullet 3 + 2^{2} \bullet 3^{2} + 2^{3} \bullet 3^{3} + ... + 2^{n} \bullet 3^{n}}}$
(as $S_{rs} = 0,\$if $r \neq s$ and $S_{rs} = 1,$ if $r = s)$
$= \frac{6(6^{n} - 1)}{6 - 1} = \frac{6}{5}(6^{n} - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.