How do you write the explicit formula for the sequence 4,8,16,32,64...?

1 Answer
Apr 4, 2016

Answer:

#a_n = 2^((n+1))#

Explanation:

Let any term in the sequence by #a#
Then the #n^(th)# term is #a_n#

First write this out:

color(white)("d")

#ncolor(white)(.) -> color(white)("d") 1color(white)("d") 2color(white)("d") 3color(white)("dd") 4color(white)("dd") 5#
#a_n->color(white)("d")4color(white)("d") 8 color(white)("d") 16color(white)("d") 32 color(white)("d") 64"#

Apart from the starting point notice that each term is #a_n = 2xx a_(n-1)#

So, for example, term 4 is #2xx" term 3"#

#ncolor(white)(.) -> color(white)("d")1color(white)("d.d") 2 color(white)("ddd") 3 color(white)("d..") 4 color(white)(d"d") 5#
#underline(a_ncolor(white)("d")->4color(white)("ddd") 8color(white)("dd") 16 color(white)("d.")32color(white)("d,") 64)#
#color(white)("dddddd")2^2color(white)(..)2^3color(white)(..)2^4color(white)(..)2^5color(white)(..)2^6#

But we need to relate the #x# in #2^x# to #n#

Notice that in each case #x=n+1#

So we have #a_n = 2^((n+1))#