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))
