Question

tính và so sánh 1.1! + 2.2! + 3.3! +...+2004.2004! và 2005!

Answer 1

Ta có: \(n\cdot n!=\left(n+1-1\right)\cdot n!=\left(n+1\right)n!-n!=\left(n+1\right)!-n!\)

(vì \(n!=1\cdot2\cdot3\cdot...\cdot n\Rightarrow\left(n+1\right)n!=1\cdot2\cdot3\cdot...\cdot n\cdot\left(n+1\right)=\left(n+1\right)!\)) 

 \(1\cdot1!+2\cdot2!+3.3!+4.4!+...+2004\cdot2004!\)

\(=2!-1!+3!-2!+4!-3!+5!-4!+...+2005!-2004!\)

\(=2005!-1!\)

\(=2005!-1\)

Mà: \(2005!-1< 2005!\)

\(\Rightarrow1\cdot1!+2\cdot2!+3\cdot3!+...+2004\cdot2004!< 2005!\)