Answer:
54
Explanation:
We are given the sequence or function:—
![\displaystyle \large{a(n) = 3(2)^n}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xx16t7bslecvtw78m2izr2egmj6pfyizyn.png)
To find the sum of 1st term and 4th term, first we must find the first term and the fourth term which can be done by substituting n = 1 and n = 4.
![\displaystyle \large{a(1)=3(2)^1}\\\displaystyle \large{a(1)=3 \cdot 2}\\\displaystyle \large{a(1)=6}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lsd6djwmjuuk08pnzwn2cmqvwj682ovlju.png)
Then when n = 4.
![\displaystyle \large{a(4) = 3(2)^4}\\\displaystyle \large{a(4) = 3 \cdot 16}\\\displaystyle \large{a(4) = 48}](https://img.qammunity.org/2023/formulas/mathematics/high-school/anlapv2bn9ftucwfn43dlfp3phey6tqwlb.png)
Therefore, the sum of 1st term and 4th term is:—
![\displaystyle \large{a(1)+a(4) = 6+48}\\\displaystyle \large{a(1)+a(4)=54}](https://img.qammunity.org/2023/formulas/mathematics/high-school/s8e0vbpvnegn0ciqdp0jdez87phtqn9yaj.png)