Question

Recursive Sequences A sequence is defined recursively by the given formulas. Find the first five terms of the sequence.

$a_n=2\left(a_{n-1}+3\right)\text{and}{a}_1=4$

Short answer

The first five terms are 4, 14, 34, 74, 154.

Step-by-step solution

Step 1.  Given

The given sequence is $a_n=2\left(a_{n-1}+3\right) \text{and}{a}_1=4$.

Step 2.  To determine

We have to find the first five terms of the given sequence.

Step 3.  Calculation

To find the first five terms of the sequence we’ll plug n = 1, 2, 3, 4 and 5 respectively in the given sequence.

$\begin{array}{l}\text{For\hspace{0.17em}\hspace{0.17em}}n=1,a_1=4\\ \text{For\hspace{0.17em}\hspace{0.17em}}n=2,a_2=2\left(a_{2-1}+3\right)=2\left(a_1+3\right)=2\left(4+3\right)=14\\ \text{For\hspace{0.17em}\hspace{0.17em}}n=3,a_3=2\left(a_{3-1}+3\right)=2\left(a_2+3\right)=2\left(14+3\right)=34\\ \text{For\hspace{0.17em}\hspace{0.17em}}n=4,a_4=2\left(a_{4-1}+3\right)=2\left(a_3+3\right)=2\left(34+3\right)=74\\ \text{For\hspace{0.17em}\hspace{0.17em}}n=5,a_5=2\left(a_{5-1}+3\right)=2\left(a_4+3\right)=2\left(74+3\right)=154\end{array}$

Hence, the first five terms are 4, 14, 34, 74, 154.