How can I find the first 5 numbers of a the sequence knowing that A(n)= 1 and A(n+1)= 4A(n)-2 ?

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Sorry if the formulas aren't very clear but everything that is between parentheses is supposed to be in subscript but I can't seem to find how to do it on this website. Thank you in advance

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Answer 1 · 2018-07-30T03:51:23

Below

$A_{n} = 1$ $A_n=1$
$A_{n + 1} = 4 A_{n} - 2$ $A_(n+1)=4A_n-2$

Therefore, if you want the first 5 numbers, then you want to find $A_{1}, A_{2}, A_{3}, A_{4} and A_{5}$ $A_1, A_2, A_3, A_4 "and" A_5$

$A_{1} = 1$ $A_1=1$

$A_{2} = 4 A_{1} - 2$ $A_2=4A_1-2$
$A_{2} = 4 \times 1 - 2$ $A_2=4times1-2$
$A_{2} = 2$ $A_2=2$

$A_{3} = 4 A_{2} - 2$ $A_3=4A_2-2$
$A_{3} = 4 \times 2 - 2$ $A_3=4times2-2$
$A_{3} = 6$ $A_3=6$

$A_{4} = 4 A_{3} - 2$ $A_4=4A_3-2$
$A_{4} = 4 \times 6 - 2$ $A_4=4times6-2$
$A_{4} = 22$ $A_4=22$

$A_{5} = 4 A_{4} - 2$ $A_5=4A_4-2$
$A_{5} = 4 \times 22 - 2$ $A_5=4times22-2$
$A_{5} = 86$ $A_5=86$