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A(1)=1, A(2)=2, A(3)= 3, A(4)=6, A(5)=12
Seq after n>3 is [2^(n-3)] *3

Let us say x=3
so A(14)= [(2^13)*x]
and A(11)= [(2^8)*x]

ratio would be [(2^13)*x] / [(2^8)*x]
then it would be just [(2^11)]/[(2^8)]= 2^3
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Can you please tell me how derived information for A3
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Can you please tell me how derived information for A3


We are given that each next term of the seq if formed by adding all the previous terms.

A3= A1+A2
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AbhimanyuDhar
A(1)=1, A(2)=2, A(3)= 3, A(4)=6, A(5)=12
Seq after n>3 is [2^(n-3)] *3

Let us say x=3
so A(14)= [(2^13)*x]
and A(11)= [(2^8)*x]

ratio would be [(2^13)*x] / [(2^8)*x]
then it would be just [(2^11)]/[(2^8)]= 2^3

Hey AbhimanyuDhar,
Thanks for the insight. That is very helpful.
I just wanted to know how we know that after n>3, [2^(n-3)] *3 holds true for every term?
Is there a principle?
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