Question

Pe 25 si 26 dar daca nu.l stit pe 26 puteti sa.l faceti pe 25(neaparat)

Si mai imi trebuie aceste ex

1.Determina numărul natural x ,astfel încât 0,(2)*+0,(3)*+0,(4)*=29/81. (* este x)

2.Fie S=1/3+1/3²+1/3³+...+1/3^49.Arata ca S<1/3


Repede....
Multumesc
Dau 30 de PUNCTE.

Answer 1

Ex.25
0,(3)^n + 0,(3)^(n+1) + 0,(3)^(n+2) = 13/81
(3/9)^n + (3/9)^(n+1) + (3/9)^(n+2) = 13/81
(1/3)^n + (1/3)^(n+1) + (1/3)^(n+2) = 13/81
(1/3)^n [1 + 1/3 + (1/3)^2] = 13/81
(1/3)^n (1+1/3+1/9) = 13/81
(1/3)^n (9/9+3/9+1/9) = 13/81
(1/3)^n x 13/9 = 13/81
1/3^n x 13/3^2 = 13/3^4
13/3^(n+2) = 13/3^4
n+2 = 4
n = 4-2
n = 2

Ex.26
1/2^2 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10 = 1/4+1/16+1/64+1/256+1/1024 = 256/1024+64/1024+16/1024+4/1024+1/1024 = 341/1024

4+4^2+4^3+4^4+4^5 = 4+16+64+256+1024=1364=3x341

[(1/2^2+1/2^4+1/2^6+1/2^8+1/2^10) / (4+4^2+4^3+4^4+4^5)] x 256 = [(341/1024) / 3x341] x 256 = (341/1024 x 1/341x3) x 256 = (1/3x1024) x 256 = 1/3x4 = 1/12 

0,(2)^x + 0,(3)^x + 0,(4)^x = 29/81
(2/9)^x + (3/9)^x + (4/9)^x = 29/81
2^x/9^x + 3^x/9^x + 4^x/9^x = 29/81
(1/9^x) (2^x+3^x+4^x) = 29/9^2
9^x = 9^2, adica x=2
Verificam : 2^2+3^2+4^2 = 4+9+16 = 29, se verifica


La celallat exercitiu, S nu poate fi mai mic decat 1/3, pentru ca primul termen al lui S este chiar 1/3, la care se mai adauga alti termeni pozitivi, deci S>1/3