[MD-sorular] öss deneme sorusu

[Can Demir]

Sorunun ifadesi biraz garip ama abladýðým kadarýyla 3^2+3^3+....+3^99+3^100=x ise 3+3^2+3^3+....+3^101+3^102=? diye soruluyor.
 
3^2+3^3+....+3^99+3^100=x ise 3(3^2+3^3+....+3^99+3^100)=3^3+3^4+....+3^100+3^101=3x olur. 3^3+3^4+....+3^100=x-(3^2) olduðundan 3^101=3x-x+(3^2)=2x+9 bulunur. Buradan da 3^102=3*(3^101)=3*(2x+9)=6x+27 eder. Bize sorulan ifade, 3+3^2+3^3+....+3^101+3^102=3+x+3^101+3^102=3+x+2x+9+6x+27=9x+39 olarak bulunur.
 

bülent aydýn <antalyacatblue at yahoo.com> wrote:
3^2+3^3+3^4+....+3^n=x olduðuna göre ,  n=100 için 3+3^2+3^3+3^4+.....+3^102= ?
('^' üssü ifade etmektedir)
 
a) 9x+3    b) 3x+12     c)9x+18       d)9x+12         e)9x+39
 
Bülent Aydýn

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