Bonjour,
1)
[tex] \forall\ n\geq 4:\ 2^n\geq 4n\\Initialisation\ n=4: 2^4=16\geq4*4\\H\'er\'edit\'e:\\On\ suppose\ la\ proposition\ vraie\ \forall\ n\geq 4:\ 2^n\geq 4n\\et\ on\ va\ d\' emontrer\que\ \forall\ n\geq 4:\ 2^{n+1}\geq 4{n+1}\\Lemme:\\n\geq1\\\Rightarrow\ 4*n\geq 4*1\\\Rightarrow\ 8n-4n\geq 4\\\Rightarrow\ 8n\geq 4n+4\\\Rightarrow\ 2*4n\geq 4(n+1).\\\\\forall\ n\geq 4:\ 2^{n+1}=2*2^n\geq 2*4n\geq4*(n+1)\\ [/tex]
2)
[tex] \forall\ n\geq \ 1:\\\dfrac{1}{1*2} +\dfrac{1}{2*3}+\dfrac{1}{3*4}+...+\dfrac{1}{n*(n+1)}=\sum_{i=1}^n \dfrac{1}{i*(i+1)}=1-\dfrac{1}{n+1}\\Initialisation:\\n=1; \dfrac{1}{1*2}=\dfrac{1}{2}=1-\dfrac{1}{2}\\H\' er\' edit\' e:\\\forall\ n\geq \ 1:\ \sum_{i=1}^n \dfrac{1}{i*(i+1)}=1-\dfrac{1}{n+1}\ est\ vrai\\alors\\\forall\ n\geq \ 1:\ \sum_{i=1}^{n+1} \dfrac{1}{i*(i+1)}=1-\dfrac{1}{n+2}\ est\ vrai\\ [/tex]
[tex] \sum_{i=1}^{n+1} \dfrac{1}{i*(i+1)}=\sum_{i=1}^{n} \dfrac{1}{i*(i+1)}+\dfrac{1}{(n+1)(n+2)} \\=1-\dfrac{1}{n+1} +\dfrac{1}{(n+1)(n+2)} \\=1-\dfrac{n+2}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)} \\=1-\dfrac{n+2-1}{(n+1)(n+2)} \\=1-\dfrac{n+1}{(n+1)(n+2)} \\=1-\dfrac{1}{n+2} \\ [/tex]
Rem:
[tex] \dfrac{1}{n}- \dfrac{1}{n+1} =\dfrac{n+1-n}{n(n+1)} =\dfrac{1}{n(n+1)}\\\dfrac{1}{1}- \dfrac{1}{2} =\dfrac{1}{1*2}\\\dfrac{1}{2}- \dfrac{1}{3} =\dfrac{1}{2*3}\\\dfrac{1}{3}- \dfrac{1}{4} =\dfrac{1}{3*4}\\...\\\dfrac{1}{n}- \dfrac{1}{n+1} =\dfrac{1}{n*(n+1)}\\\\on\ additionne\ \membre \`a\ membre\\1-\dfrac{1}{n+1}=\sum_{i=1}^n\ \dfrac{1}{i*(i+1)}\\ [/tex]
