Question

Write in set builder form A={1,5,19,65,211}

Answer 1

Answer:

$A=\left\{x|x_n=3\cdot x_{n-1}+2^n:x_0=1\:\:,for\:n=0,1,2,3,4\right\}$

Step-by-step explanation:

Here we need to find

This kind of recurrence tends to happen with exponential-type series.

$5=1*3+2,\\19=5*3+22,\\65=19*3+23,\\211=65*3+24$

You can find them by plugging in the first couple of terms in your series, thus producing

IF $a_0=1 ,\ then  \ a_n=3*a_{n-1}+2^n$

thus we can write it as

$A=\left\{x|x_n=3\cdot x_{n-1}+2^n:x_0=1\:\:,for\:n=0,1,2,3,4\right\}$