Question

$Fie~nr.~a= \frac{1}{ 2^{2}} + \frac{1}{ 3^{2}} + \frac{1}{ 4^{2} } +...+ \frac{1}{ 100^{2} } .~Aratati~ca~0,2< \sqrt{ \frac{a}{11} }<0,3.$

Answer 1

Fie n∈N*.

$\frac{1}{n(n+1)} <\frac{1}{ n^{2} }< \frac{1}{(n-1)n}$

Astfel: $\frac{1}{2^{2} } < \frac{1}{1*2} \\ \frac{1}{ 3^{2} } < \frac{1}{2*3} \\ .............. \\ \frac{1}{100^{2} } < \frac{1}{99*100}$

Insumand, obtinem: $a< \frac{1}{1*2}+ \frac{1}{2*3}+...+ \frac{1}{99*100} = \frac{99}{100} ~(SUMA~TELESCOPICA)$. Asadar $\frac{a}{11}< \frac{ \frac{99}{100} }{11} = \frac{9}{100} => \sqrt{ \frac{a}{11} }< \sqrt{ \frac{9}{100} } =0,3$......... (1)

Pentru a demonstra "cealalta parte", procedam analog:

$\frac{1}{ 2^{2} }> \frac{1}{2*3} \\ \frac{1}{ 3^{2} }> \frac{1}{3*4} \\............. \\ \frac{1}{ 100^{2} }> \frac{1}{100*101}$. Insumand, obtinem a>$\frac{1}{2*3}+ \frac{1}{3*4}+...+ \frac{1}{100*101} = \frac{1}{2}- \frac{1}{101}= \frac{99}{202} (SUMA~TELESCOPICA)$. Deci $\frac{a}{11} > \frac{ \frac{99}{202} }{11}= \frac{9}{202} => \sqrt{ \frac{a}{11}}> \sqrt{ \frac{9}{202}} = \frac{3 \sqrt{202} }{202} > 0,2$........... (2)

*$\frac{3 \sqrt{202} }{202}>0,2$ se verifica eventual prin calcul, prin ridicare la patrat sau prin verificarea produsului dintre mezi si extremi.

Din (1) si (2) => $0,2< \sqrt{ \frac{a}{11} } <0.3$. (Sau cum imi place mie sa spun: $\sqrt{ \frac{a}{11} }$∈(0,2 ; 0,3).