Question

The value of the sum $\sum _{r=1}^n\sum _{s=1}^n{S}_{rs}{2}^r{3}^s$ where $S_{rs}=0$ if $r\ne s$ and $S_{rs}=1$ if $r=s$ is

• A. $\frac{(5^n-1)}{4}$
• B. $\frac{6}{5}(6^n-1)$
• C. $\frac{5^n{6}^n}{n+1}$
• D. $\frac{5}{4}(5^n-1)$

Answer 1

The correct option is B $\frac{6}{5}(6^n-1)$

${\sum }_{r=1}^n{\sum }_{S=1}^n{S}_{rS}{2}^r{3}^S$

$S_{rS}=0,r\ne S$

$S_{rS}=1,r=S$

$S_{11}{2}^1{3}^1+S_{22}{2}^2{3}^2+S_{33}{2}^3{3}^3+....+S_{nn}{2}^n{3}^n$

$=2^1{3}^1+2^2{3}^2+2^3{3}^3+...2^n{3}^n$

$=6+6^2+6^3+...+6^n$

$=\frac{6{(6}^n-1)}{6-1}=\frac{6}{5}(6^n-1)$