Question

Solve the recurrence relation \(a_n = 2^na_{n-1}\) with \(a_0 = 1\)

Answer 1

I try \(a_n = 2^n a_{n-1} = 2^n(2^{n-1}a_{n-2}) = 2^{2n-1}a_{n-2} \)

Is that a step in the right direction?

Answer 2

If we continue your equalities, we will get:
\[a _{n}=2^{n}a _{n-1}=2^{n}2^{n-1}a _{n-2}=...2^{n}2^{n-1}2^{n-2}...2^{1}a _{0}\]

Answer 3

so \[a _{n}=2^{(1+n)\frac{ n }{ 2 }}\]

Answer 4

the formula above makes sense, because:
if n=0, then a_0=1
if n=1, then a_1=2
if n=3, then a_3=2^6=64
and so on...
@calyne

Answer 5

amazing thank you

Answer 6

thank you!