Question

[tex]Demonstrati\ urmatoarele\ egalitati:\\
a)\frac{(n+1)(n+2)*...*(n+n)}{1*3*5*...*(2n-1)}=2^n, \vee\ n\in N^*\\
b)[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{n^2}]=\frac{n(4n^2-3n+5)}{6},\ \vee\ n \in N^*\\
c)1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}, \vee n \in N^*[/tex]

Answer 1

$a)~i)~Pentru~n=1~avem:~ \frac{2}{1}=2^1,~adevarat. \\ \\ ii)~Clasica~presupunere~(p(k)-adevarat,~k \in N^*). \\ \\ p(k+1)~:~ \frac{(k+2)(k+3)...(k+k)}{1 \cdot 3 \cdot 5 \cdot ... \cdot (2k-1)} \cdot \frac{(2k+1)(2k+2)}{2k+1}= \frac{2^k}{k+1} \cdot 2(k+1)=2^{k+1}. \\ \\ b)~i)~Pentru~n=1~avem~:~ [ \sqrt{1}]= \frac{1(4 \cdot 1^2-3 \cdot 1+5)}{6},~adevarat. \\ \\ ii)~Clasica~presupunere~(p(k)-adevarat,~k \in N^*). \\ \\ Voi~nota~S_n=[ \sqrt{1}]+ [ \sqrt{2}]+...+[ \sqrt{n^2} ].$

$S_{k+1}=S_k+ [ \sqrt{k^2+1}]+ [\sqrt{k^2+2}]+...+ [ \sqrt{k^2+2k}]+ [\sqrt{(k+1)^2}]= \\ \\ = \frac{k(4k^2-3k+5)}{6}+2k \cdot k+k+1= \frac{4k^3+9k^2+11k+6}{6}= \\ \\ = \frac{(k+1)(4(k+1)^2-3(k+1)+5)}{6} ~(aceasta~ultima~egalitate~se~verifica~fie \\ \\ prin ~calcul,~fie~prin~descompunere:~(4k^3+4k^2)+(5k^2+5k)+ \\ \\ +(6k+6)).$

$\displaystyle c)~i)~Pentru~n=1~avem~1- \frac{1}{2} = \frac{1}{2},~adevarat. \\ \\ ii)~Clasica~presupunere~(p(k)-adevarat~;~ k \in N^*). \\ \\ Notez~S_{n}=1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+...+ \frac{1}{2n-1}- \frac{1}{2n}. \\ \\ S_{k+1}=S_k+ \frac{1}{2k+1}- \frac{1}{2k+2}= \\ \\ = \boxed{\frac{1}{k+1}}+ \frac{1}{k+2}+...+ \frac{1}{2k}+ \frac{1}{2k+1} \boxed{- \frac{1}{2k+2}}= \\ \\ = \frac{1}{k+2}+ \frac{1}{k+3}+...+ \frac{1}{2(k+1)},~q.e.d.$