Solution
- Let us evaluate the summations given to us and we can thus know the order.
[tex]\begin{gathered} \sum ^5_{k=1}4(2)^{k-1}=4(2)^0+4(2)^1+4(2)^2+4(2)^3+4(2)^4 \\ \\ \sum ^5_{k=1}4(2)^{k-1}=4+8+16+32+64=124 \end{gathered}[/tex]
[tex]\begin{gathered} \sum ^4_{k=1}3(4)^{k-1}=3(4)^0+3(4)^1+3(4)^2+3(4)^3 \\ \\ \sum ^4_{k=1}3(4)^{k-1}=3+12+48+192=255 \end{gathered}[/tex]
[tex]\sum ^3_{k=1}6^{k-1}=6^0+6^1+6^2=1+6+36=43[/tex]
[tex]\begin{gathered} \sum ^2_{k=1}2(5)^{k-1}=2(5)^0+2(5)^1 \\ \\ \sum ^2_{k=1}2(5)^{k-1}=2+10=12 \end{gathered}[/tex]
- Therefore, we can order the sums
