[tex]a)~ \frac{k}{(k+1)!}= \frac{(k+1)-1}{(k+1)!}= \frac{1}{k!}- \frac{1}{(k+1)!} . \\ \\ \sum\limits^n_{k=1} \frac{k}{(k+1)!}= \sum\limits^n_{k=1} \Big( \frac{1}{k!}- \frac{1}{(k+1)!} \Big)= \frac{1}{1!}- \frac{1}{2!}+ \frac{1}{2!}- \frac{1}{3!}-+...+ \frac{1}{n!}- \frac{1}{(n+1)!}= \\ \\ =1- \frac{1}{(n+1)!} . \\ \\ Demonstratie~prin~inductie:~ \\ \\ i)~n=1~adevarat~( \frac{1}{2!}= 1-\frac{1}{2!}). [/tex]
[tex]ii)~Presupunem~ca~propozitia~este~adevarata~pentru~n=t \in N^*~si~ \\ \\ demonstram~ca~este~adevarata~si~pentru~n=t+1. \\ \\ \sum\limits^{n+1}_{k=1} \frac{k}{(k+1)!}= \sum\limits^n_{k=1} \frac{k}{(k+1)!}+ \frac{n+1}{(n+2)!}= 1- \frac{1}{(n+1)!}+ \frac{n+1}{(n+2)!}= 1- \frac{n+2}{(n+2)!} + \\ \\ + \frac{n+1}{(n+2)!} =1- \frac{1}{(n+2)!}. [/tex]
[tex]b)~k \cdot k!=((k+1)-1) k!=(k+1)k!-k!=(k+1)!-k!. \\ \\ S_n=2!-1!+3!-2!+4!-3!+...+(n+1)!-n!=-1+(n+1)! \\ \\ Deci~S_n=(n+1)!-1. \\ \\ Demonstratie~prin~inductie: \\ \\ i)~S_1=1 \cdot 1!=2!-1,~adevarat. \\ \\ ii)~Clasica~presupunere... \\ \\ S_{n+1}=S_n+(n+1) \cdot (n+1)!=(n+1)!-1+(n+1)(n+1)!= \\ \\ =(n+1)!(n+1+1)-1=(n+2)!-1. [/tex]
[tex]c)~S_n=1+(2+2^2+2^3+...+2^{n-1}+2^n) \\ \\ ~~~2S_n=(2 + 2^2+2^3+2^4+...+2^n)+2^{n+1} \\ ---------------- \\ ~~~2S_n-S_n=2^{n+1}-1 \Leftrightarrow S_n = 2^{n+1}-1. \\ \\ Demosntratie~prin~inductie: \\ \\ i)~S_1=1+2=2^{1+1}-1~(adevarat). \\ \\ ii)~Clasica~presupunere... \\ \\ S_{n+1}=S_n+2^{n+1}=2^{n+1}+2^{n+1}-1=2 \cdot 2^{n+1}-1=2^{n+2}-1...gata![/tex]
