EXERCITIUL 8.va rog frumos ajutatima!

aducem la acelasi numitor 1-1/4=(4-1)/4=3/4,1-1/9=(9-1)/9=8/9/,1-1/n patrat=npatrat-1/n la patrat
Pn=3/4x8/9-npatrat-1/npatrat
Pn=24/36-npatrat-1/npatrat
Pn=24npatrat-36(n la patrat-1)=24n patrat-36 n patrat+36=-12npatrat+36

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Emil1234 Emil1234 · Answer 2 · 2014-11-01T14:29:44+02:00

8.

[tex] P_{n}= (1- \frac{1}{4})(1- \frac{1}{9})....(1- \frac{1}{ n^{2} }) \\ 1- \frac{1}{ n^{2} } = \frac{ n^{2}-1 }{ n^{2} } = \frac{(n-1)(n+1)}{ n^{2} } \\ \\ P_{n} = \frac{(2-1)(2+1)}{4} \frac{(3-1)(3+1)}{9}...... \frac{(n-1)(n+1)}{ n^{2} } \\ P_{n} = \frac{1*3}{4} \frac{2*4}{9}...... \frac{(n-2)n}{ (n-1)^{2} } \frac{(n-1)(n+1)}{ n^{2} } [/tex]

Se observa factul ca se simplifica termenii . Vom ajunge la urmatoarea formula :

[tex] P_{n} = \frac{1}{2} \frac{(n+1)}{n} = \frac{n+1}{2n} [/tex]