g is defined recursively by g(1)=2 and g(n)=2g(n-1) where n is greater than/equal to 2...solve for g(4)

Question

that is g(n)=2^g(n-1)

 

i think i am stuck on where to plug in what....

i know that a series/sequence defined recursively is g(n)=g⁽ⁿ⁺¹⁾=(n+1)g(n)

so if g(1)=2 then for the g(n)=2^g(n-1)  for n≥2  then g(1)=2⁰=1

 

then for g(2) im kinda lost on what to plug in where, is it 2^2(3-1)?  that gives 16

 

 

i know that in the end the answer will be g(4)=4*g(3) but i cant seem to get g(2) and g(3)

Answer 1

g(2) = 2^g(1) = 2² = 4

 

g(3) = 2^g(2) = 2⁴ = 16

 

g(4) = 2^g(3) = 2¹⁶ = 65,536

Answer 2

g(1)=2.  Now g(n) is given by 2^(g(n-1))

 

for all n >= 2.  This means that g(2) = 2^(g(2-1)) = 2^(g(1)) = 2^2 = 4

 

g(3) = 2^(g(3-1)) = 2^(g(2)) = 2^4 = 16

 

g(4) = 2^(g3) = 2^16 which is the answer.