Forgot password? New user? Sign up
Existing user? Log in
We define a recursive sequence of natural numbers such that a1=2a_{1}=2a1=2 and an+1=3an+2a_{n+1}=3a_{n}+2an+1=3an+2 for n≥1n\geq1n≥1. Find [2∑n=1100an]+200a100\frac{[2\sum_{n=1}^{100}a_{n}]+200}{a_{100}}a100[2∑n=1100an]+200.
The problem is original.
Problem Loading...
Note Loading...
Set Loading...