Question

Puteti sa ma ajutati la 18!?

Answer 1

$E(n)=n^3+(n-3)(n+3)-(n+1)(n^2-n-1)+9=n^3+n^2-9-(n^3-n^2-n+n^2-n-1)+9=n^3+n^2-(n^3-2n-1)=n^3+n^2-n^3+2n+1=n^2+2n+1=(n+1)^2,\forall n \in\mathbb{Z}$

Answer 2

$\it E(n)=n^3+(n-3)(n+3)-(n+1)(n^2-n-1)+9 \Rightarrow \\ \\ \Rightarrow E(n) = n^3+n^2-\not9-(n+1)(n^2-n-1)+\not9 \Longrightarrow \\ \\ \Rightarrow E(n)=n^2(n+1)-(n+1)(n^2-n-1)\Rightarrow E(n)=(n+1)(\not n^2-\not n^2+n+1)\Rightarrow\\ \\ \Rightarrow E(n) = (n+1)(n+1)=(n+1)^2$