Question

Să se verifice egalitățile:

a) 2!/1! + 3!/2! + ... + (n+1)!/n! = n x (n+3)/2, n aparține N*
b) 2!/0! + 3!/1! + ... + (n+1)!/(n-1)! = n(n+1)(n+2)/3

Answer 1

$a).\displaystyle \frac{2!}{1!} + \frac{3!}{2!} +...+ \frac{(n+1)!}{n!} = \frac{n (n+3)}{2} \\ \\ \frac{2!}{1!} + \frac{3!}{2!} +...+ \frac{(n+1)!}{n!} = \frac{\not1 \cdot 2}{\not1} + \frac{\not1 \cdot \not2 \cdot 3}{\not1 \cdot \not2} +...+n+1= \\ \\ =2+3+...+(n+1)= \Sigma_{k=1}^n(k+1)=\Sigma_{k=1}^nk+\Sigma_{k=1}^n1= \\ \\ = \frac{n(n+1)}{2} +n= \frac{n(n+1)+2n}{2} = \frac{n^2+n+2n}{2} = \frac{n^2+3n}{2} = \frac{n (n+3)}{2}$
[tex]\displaystyle \frac{2!}{1!} + \frac{3!}{2!} +...+ \frac{(n+1)!}{n!} = \frac{n(n+3)}{2} \Rightarrow \frac{n(n+3)}{2} = \frac{n(n+3)}{2} ~(A)[/tex]

[tex]\displaystyle b). \frac{2!}{0!} + \frac{3!}{1!} +...+ \frac{(n+1)!}{(n-1)!} = \frac{n(n+1)(n+2)}{3} \\ \\ \frac{2!}{0!} + \frac{3!}{1!} +...+ \frac{(n+1)!}{(n-1)!} = \frac{\not1 \cdot 2}{\not1} + \frac{\not1 \cdot 2 \cdot 3}{\not1} +...+n(n+1)= \\ \\ =2+2 \cdot 3+...+n(n+1)=\Sigma _{k=1}^n[k(k+1)]=\Sigma_{k=1}^n(k^2+k)= \\ \\ [/tex] [tex]\displaystyle =\Sigma_{k=1}^nk^2+\Sigma _{k=1}^nk= \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \\ \\ = \frac{(n^2+n)(2n+1)}{6} + \frac{n^2+n}{2} = \frac{2n^3+n^2+2n^2+n}{6} + \frac{n^2+n}{2} = \\ \\ = \frac{2n^3+n^2+2n^2+n+3(n^2+n)}{6} = \frac{2n^3+n^2+2n^2+n+3n^2+3n}{6} = \\ \\ = \frac{2n^3+6n^2+4n}{6} = \frac{\not2(n^3+3n^2+2n)}{\not6} = \frac{n^3+3n^2+2n}{3} = \\ \\ = \frac{n(n^2+3n+2)}{3} = \frac{n(n^2+n+2n+2)}{3} = \frac{n[n(n+1)+2(n+1)]}{3} = \\ \\ = \frac{n(n+1)(n+2)}{3} [/tex]
$\displaystyle \frac{2!}{0!} + \frac{3!}{1!} +...+ \frac{(n+1)!}{(n-1)!} = \frac{n(n+1)(n+2)}{3} \Rightarrow \\ \\ \Rightarrow \frac{n(n+1)(n+2)}{3} = \frac{n(n+1)(n+2)}{3}~(A)$