Question

Calculati:$\frac{5}{3}$+$\frac{32}{15}$+$\frac{107}{35}$+...+$\frac{31982}{1599}$-$\frac{40}{41}$
Va rog!

Answer 1

prima data incerci sa descoperi cate fractii sunt(de la 5/3 la 31982/1599):
prima fractie ai la numitor 1*3...la a doua ai 3*5,la a treia 5*7...la ultima 39*41

observi ca ai 1,3,5,....39...ai numere din 2 in 2, incepand cu 1 pana la 39.

Nr de termeni = (ultimul termen-primul termen)/ratie +1 (unde ratia, in cazul tau, este 2(nr cresc din 2 in 2 ) )
=> ai (39-1):2 +1 =20 de fractii
revenim la suma:
[tex] \frac{3}{5} + \frac{32}{15} + \frac{107}{35} ...+ \frac{31982}{1599} = \\ =( \frac{3}{3} + \frac{2}{3} )+ (\frac{15*2}{15} + \frac{2}{15} )+( \frac{3*35}{35}+ \frac{2}{35} )+...+( \frac{20*1599}{1599} + \frac{2}{1599} )= \\ =(1+2+3+...+20)+( \frac{2}{3} + \frac{2}{15}+ \frac{2}{35}+ \frac{2}{1599} )= \\ =210+2*( \frac{1}{3}+ \frac{1}{15} +\frac{1}{35} + \frac{1}{1599} ) [/tex]

Acum ai o suma telescopica :
[tex] \frac{1}{1*3}= \frac{1}{2}*( \frac{1}{1} -\frac{1}{3} ) \\ \frac{1}{3*5}= \frac{1}{2} *( \frac{1}{3} - \frac{1}{5} ) \\ ............................. \\ \frac{1}{39*41} = \frac{1}{2} *( \frac{1}{39} - \frac{1}{41} ) \\ ============ \\ S= \frac{1}{2} *( \frac{1}{1} - \frac{1}{41} )= \frac{1}{2} * \frac{40}{41} \\ revenind: 210+2* \frac{1}{2} * \frac{40}{41}- \frac{40}{41} = \\ =210[/tex]