nick_sun wrote:
prude_sb wrote:
a(n) = nth element in the sequence
this means that a(n-1) - a(n) = a(n) => a(n-1) = 2*a(n) => a(n) = a(n-1)/2
so if we go on substituting this formula until n = 1 on the RHS we get
a(n) = a(1)/2^(n-1)
so a(4) = a(1)/2^3
so either stat1 or stat2 can be used to get the ans
my ans is D
Could anybody please explain how Prude has obtained powers in the equation marked above?
Once we have a(n) = a(n-1)/2, we can calculate the relation in term n-2, n-3.... 1.
Notice that it's geometry sequence with r=1/2. Formulas helps us to conclude directly.
This is a full detailed way to arrive at the answer
a(n)
= a(n-1)/2 >>> with n-1
= 1/2 * (a(n-2)/2) >>> with n-2
= 1/2 * 1/2 * (a(n-3)/2) >>> with n-3
= (1/2)^3 * a(n-3)
= (1/2)^p * a(n-p) >>> with n-p
Then, if p = n-1, we have:
a(n)
= (1/2)^p * a(n-p)
= (1/2)^(n-1) * a(n-(n-1))
= (1/2)^(n-1) * a(1)