The required explicit formula is ![a_n = -3(a_{n - 1})](https://tex.z-dn.net/?f=a_n%20%3D%20-3%28a_%7Bn%20-%201%7D%29)
<h3><u>Solution:</u></h3>
Given that sequence is -4, 12, -36, 108
<em><u>To find: explicit formula</u></em>
Explicit formulas define each term in a sequence directly, allowing one to calculate any term in the sequence
An explicit formula designates the nth term of the sequence, as an expression of n (where n = the term's location). It defines the sequence as a formula in terms of n.
<em><u>Let us first find the logic used in sequence</u></em>
![\begin{array}{l}{-4 \times-3=12} \\\\ {12 \times-3=-36} \\\\ {-36 \times-3=108}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B-4%20%5Ctimes-3%3D12%7D%20%5C%5C%5C%5C%20%7B12%20%5Ctimes-3%3D-36%7D%20%5C%5C%5C%5C%20%7B-36%20%5Ctimes-3%3D108%7D%5Cend%7Barray%7D)
So we can see clearly that next term in sequence is obtained by multiplying -3 with previous term
This can be defined in terms of "n"
![a_n = -3(a_{n - 1})](https://tex.z-dn.net/?f=a_n%20%3D%20-3%28a_%7Bn%20-%201%7D%29)
Where
represents the next terms location and
represents previous term location
So the required explicit formula is ![a_n = -3(a_{n - 1})](https://tex.z-dn.net/?f=a_n%20%3D%20-3%28a_%7Bn%20-%201%7D%29)
<em><u>Let us verify our explicit formula</u></em>
Now let us find the 4th term of sequence
![\text {so } a_{n}=a_{4} \text { and } a_{n-1}=a_{4-1}=a_{3}](https://tex.z-dn.net/?f=%5Ctext%20%7Bso%20%7D%20a_%7Bn%7D%3Da_%7B4%7D%20%5Ctext%20%7B%20and%20%7D%20a_%7Bn-1%7D%3Da_%7B4-1%7D%3Da_%7B3%7D)
![a_{4}=-3\left(a_{3}\right)=-3(-36)=108](https://tex.z-dn.net/?f=a_%7B4%7D%3D-3%5Cleft%28a_%7B3%7D%5Cright%29%3D-3%28-36%29%3D108)
Thus using the explicit formula, next terms in sequence can be found