Answer to Question #317531 in Discrete Mathematics for Kiran

Question

An=2*n+an-1,A1=1

Answer 1

$a_n=2n+a_{n-1},a_1=1\\a_n=2n+a_{n-1}=2n+2\left( n-1 \right) +a_{n-2}=...=\\=2n+2\left( n-1 \right) +...+2\cdot 2+a_1=\\=1+2\left( 2+3+...+n \right) =1+2\left( \frac{n\left( n-1 \right)}{2}-1 \right) =\\=1+n^2-n-2=n^2-n-1$