Question

Suppose that a sequence id defined as follows.a1=-4,an=-2a(n-1)+6 for n>=2List the first four terms of the sequence.

Answer 1

Answer:

-4, 14, -22 and 50.

Explanation:

Given the sequence defined as follows:

$\begin{gathered} a_1=-4 \\ a_n=-2a_{n-1}+6 \end{gathered}$$\begin{gathered} \text{The second term:} \\ a_2=-2a_{2-1}+6 \\ =-2a_1+6 \\ =-2(-4)+6 \\ =8+6 \\ a_2=14 \end{gathered}$

Similarly:

$\begin{gathered} \text{The third term:} \\ a_3=-2a_{3-1}+6 \\ =-2a_2+6 \\ =-2(14)+6 \\ =-28+6 \\ a_3=-22 \end{gathered}$

Finally, the fourth term:

$\begin{gathered} a_4=-2a_{4-1}+6 \\ =-2a_3+6 \\ =-2\mleft(-22\mright)+6 \\ =44+6 \\ a_4=50 \end{gathered}$

Therefore, the first four terms of the sequence are -4, 14, -22, and 50.