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CLAUDIUCLAUDIU7ZE
a fost răspuns

Sa se demonstreze egalitatile:
a) [tex]6+24+60+..+n(n^{2} -1)=\frac{n(n+1)(n^{2}+n-2)}{4} ;n\ \textgreater \ =2[/tex]
b) ([tex]1-\frac{1}{n^{2}})[1-\frac{1}{(n+1)^{2}}]...[1-\frac{1}{(2n-1)^{2}} ]=\frac{2n-2}{2n-1} ;n -apartine N*[/tex]

Răspuns :

NICUMAVROZE
b)
1-1/n^2=(n^2-1)/ n^2= (n-1)(n+1)/n^2
1-1/(n+1)^2=n(n+2)/ (n+1)^2
...
1-1/(2n-1)^2=[(2n-2)*2n] / 2n-1)^2
P=(n-1)(n+1)/n^2 * n(n+2)/ (n+1)^2 *...*[(2n-2)*2n] / 2n-1)^2 si dupa simplificari
P=(n-1)*2 / (2n-1)=(2n-2)/ (2n-1)

a) se rezolva prin inductie matematica
etapa 1: verificam pentru n=2
6=2*3*(4+2-2) /4   adica 6=6
etapa2:
Presupunem relatia adevarata pentru n si demonstram ca e adevarata si pentru n+1
notam Sn=6+24+...+n(n^2-1)=n(n+1)(n^2+n-2)/4
atunci trebuie sa aratam ca S(n+1)=6+24+...+n(n^2-1)+(n+1)[(n+1)^2-1]= (n+1)(n+2)[(n+1)^2+(n+1)-2) / 4
adica
n(n+1)(n^2+n-2)/4 + (n+1)[(n+1)^2-1] = (n+1)(n+2)[(n+1)^2+(n+1)-2) / 4
dupa cateva simplificari si calcule simple ajungem la
n^2+5n+6=n^2+5n+6 relatie adevarata
Concluzie: presupunerea facuta este adevarata si formula sumei este cea din enunt

RAYZENZE
b) Demonstram prin inductie matematica:

[tex]\Big(1-\dfrac{1}{n^2}\Big)\Big(1-\dfrac{1}{(n+1)^2}\Big)\cdot ...\cdot \Big(1-\dfrac{1}{(2n-1)^2}\Big)=\dfrac{2n-2}{2n-1} \\ \\ \begin{array}{rcl}P(1) &=& \dfrac{0}{1} = 0 \\ \\ &=& 1-\dfrac{1}{1} = 0 \end{array}\quad (A) \\ \\ \\ \begin{array}{rcl}P(k) &=& \dfrac{2k-2}{2k-1} \\ \\ &=& \Big(1-\dfrac{1}{k^2}\Big)\Big(1-\dfrac{1}{(k+1)^2}\Big)\cdot ...\cdot \Big(1-\dfrac{1}{\Big(k+(k-1)\Big)^2}\Big) \end{array} (A)[/tex]


[tex]P(k+1) = \dfrac{2(k+1)-2}{2(k+1)-1} = \boxed{\dfrac{2k}{2k+1}}\\ \\P(k+1) = \\ \\ =\Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot \Big(1- \dfrac{1}{\Big((k+1)+(k+1-1)\Big)^2}\Big) \\ \\ = \Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot \Big(1- \dfrac{1}{\Big((k+1)+k\Big) ^2}\Big)[/tex]

[tex]= \Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot\Big(1-\dfrac{1}{\Big((k+1)+(k-1)\Big)^2}\Big)\cdot \\ \\ \cdot \Big(1- \dfrac{1}{\Big((k+1)+k\Big) ^2}\Big) \\ \\ = \Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot \Big(1-\dfrac{1}{\Big((k+1)+(k-2)\Big)^2}\Big) \cdot \\ \\ \cdot \Big(1-\dfrac{1}{\Big((k+1)+(k-1)\Big)^2}\Big) \cdot \Big(1- \dfrac{1}{\Big((k+1)+k\Big) ^2}\Big)[/tex]

[tex]= \Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot \Big(1-\dfrac{1}{(2k-1)^2}\Big)\cdot \\ \\ \cdot \Big(1-\dfrac{1}{(2k)^2}\Big) \cdot \Big(1- \dfrac{1}{(2k+1) ^2}\Big) = \\ \\ = \Big(1-\dfrac{1}{k^2}\Big)\cdot \Big(1-\dfrac{1}{(k+1)^2}\Big)\Big(1-\dfrac{1}{(k+2)^2}\Big)\cdot ...\cdot \\ \\ \cdot \Big(1-\dfrac{1}{(2k-1)^2}\Big)\cdot \Big(1-\dfrac{1}{(2k)^2}\Big) \cdot \Big(1- \dfrac{1}{(2k+1) ^2}\Big)\cdot \dfrac{1}{\Big(1-\dfrac{1}{k^2}\Big)}[/tex]

[tex]=P(k)\cdot \Big(1-\dfrac{1}{(2k)^2}\Big) \cdot \Big(1- \dfrac{1}{(2k+1) ^2}\Big)\cdot \dfrac{1}{1-\dfrac{1}{k^2}} \\ \\ = \dfrac{2k-2}{2k-1} \cdot \dfrac{(2k)^2-1}{(2k)^2}\cdot \dfrac{(2k+1)^2-1}{(2k+1)^2}\cdot \dfrac{k^2}{k^2-1} \\ \\ = \dfrac{(2k-2)(2k-1)(2k+1)(2k+1-1)(2k+1+1)\cdot k^2}{(2k-1)(2k)^2(2k+1)^2(k-1)(k+1)} \\ \\ = \dfrac{(2k-2)(2k-1)(2k+1)(2k)(2k+2)\cdot k^2}{(2k-1)(2k)^2(2k+1)^2(k-1)(k+1)} [/tex]

[tex]= \dfrac{2\cdot 2\cdot 2\cdot (k-1)(2k-1)(2k+1)(k)(k+1)\cdot k^2}{2^2\cdot (2k-1)(k)^2(2k+1)^2(k-1)(k+1)} \\ \\ = \dfrac{2\cdot k}{2k+1} \\ \\ = \boxed{\dfrac{2k}{2k+1}}\quad q.e.d.[/tex]
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