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### FRACTII

#### 1).  Demonstrati ca  1/n(n+1) = 1/n  -  1/(n+1)    , oricare ar fi n numar natural diferit de 0

Aducem la acelasi numitor termenii din partea dreapta

1/n -1/(n+1)=(n+1-n)/n(n+1)=1/n(n+1)

2). Calculati suma : 1/(1*2)  + 1/(2*3) + 1/(3*4) + .......+ 1/(9*10) , folosind formula de mai sus.

Aplicam formula pentru fiecare termen in parte

1/1*2 =1/1 - 1/2

1/2*3 =1/2 - 1/3

1/3*4 =1/3 - 1/4

.......

1/9*10=1/9 -1/10  ,  adunand termenii obtinem

(1/1 -1/2 )+(1/2 -1/3) +(1/3 -1/4) +(1/4 -1/5)+(1/5-1/6) +......+(1/9-1/10)

se observa ca se pot reduce unii termeni , si vom obtine 1/1 - 1/10=1 - 1/10=9/10=0.9

3).Calculati suma :S=1/1*2 + +1/2*3 + 1/3*4 + .........+1/2001*2002 si apoi dovediti ca

1/22 + 1/32 + 1/42 + .......+ 1/20022 < 2001/2002

Procedam la fal ca mai sus , si obtinem:

(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) +.........+(1/2001/ - 1/2002) = 1/1 - 1/2002 = (2002-1)/2002 = 2001/2002

Pentru partea a doua a exercitiului se observa ca

1/22<1/1*2

1/32<1/2*3

...............

1/20022<1/20001*20002

deci suma termenilor este mai mica decat 2001/2002

4). Demonstrati ca 1/(n(n+2)) = (1/2)*[(1/n  -  1/(n+2)] , pentru orice numar natural diferit de 0

aducem la acelasi numitor termenii din partea dreapta

(1/2)*[(n+2-n)/n(n+2)]=(1/2)*[2/n(n+2)]=1/n(n+2)

5).  Calculati   suma : 1/(1*3)  + 1/(3*5) + 1/(5*7) + ....... +1/(9*11) , folosind formula de la punctul 3

Aplicam formula pentru fiecare termen in parte

1/1*3 = (1/2)*(1/1 - 1/3)

1/3*5 = (1/2)*(1/3 - 1/5)

1/5*7 = (1/2)*(1/5 - 1/7)

1/7*9 = (1/2)*(1/7 - 1/9)

1/9*11 = (1/2)*(1/9 - 1/11)  , adunand termenii obtinem

(1/2)*[(1/1 - 1/3)+(1/3 - 1/5)+(1/5 - 1/7)+(1/7 - 1/9)+(1/9 - 1/11)

se observa ca se pot reduce unii termeni , si vom obtine (1/2)*(1/1 - 1/11)=(1/2)*(1 - 1/11)=(1/2)*(10/11)=5/11

6).  Demonstrati ca :1/3 + 1/15 + 1/35 +.......+ 1/n(n+2) =(n+1)/(2n+4)

Scriem fiecare termen al sumei sub urmatoarea forma

1/3 = 1/1*3

1/15 = 1/3*5

1/35 = 1/5*7

...............

1/n(n+2)      adunand termenii obtinem , conform formulei de mai sus (de la exercitiul 4.) :

(1/2)*[(1/1 - 1/3)+(1/3 - 1/5)+(1/5 - 1/7)+ ..........+(1/n - 1/(n+2))]

se observa ca se pot reduce unii termeni , si vom obtine  (1/2)*[1/1 - 1/(n+2)]=(1/2)*[1 - 1/(n+2)]=(1/2)*[(n+2-1)/(n+2)]=

=(1/2)*(n+1)/(n+2)=(n+1)((2n+4)

7).  Aratati ca: 1*2/0.2 + 2*3/0.3 +  3*4/0.4 + ....... + 99*100/10 =49500

vom scrie fiecare termen al sumei sub forma urmatoare

1*2/0.2 = 1*2*10/2 = 1*10=10   , stiind ca 0.2=2/10

2*3/0.2 = 2*3*10/3 = 2*10

3*4/0.4 = 3*4*10/4 = 3*10

.......................

99*100/10 = 99*100*10/100   , stiind ca 10=100/10

8).  Calculati  (0.1+0.2+0.3+.....+0.9+1):5.5

vom scrie fiecare termen al sumei sub forma urmatoare

0.1=1/10

0.2=2/10

0.3=3/10

.............

0.9=9/10

1=10/10

adunand termenii obtinem  (1/10 + 2/10 + 3/10 + ....... + 9/10+10/10):55/10     , stiind ca 5.5=55/10

vom da factor comun pe 1/10 si obtinem (1/10)*(1+2+3+........+10):55/10=(1/10)*(11*10:2):55/10=1

9).  Sa se afle x din urmatoarea ecuatie:

[(x+1/2) + (x+1/3) + (x+1/6)]:0.8=x+11

(x+1/2) + (x+1/3) + (x+1/6)=0.8*(x+11)

3x+1/2 + 1/3 +1/6 =0.8*(x+11)

3x+(3+2+1)/6 = 0.8*(x+11)   , stiind ca   0.8=8/10 vom scrie :

3x+1 =8*(x+11)/10  >  3x+1 = (8x+88)/10  >  10*(3x+1)=8x+88  >  30x+10=8x+88

30x-8x=88-10  >   22x=78  >  x=78:22=39/11

10).  Daca a = 1/2 + 2/3 + 3/4 + ........+ 99/100  si

b = 1/2 + 1/3 + 1/4 + .........+ 1/100  , atunci calculati  a+b

Adunam termenii numerelor  1/2 + 1/2 + 2/3 + 1/3 + 3/4 + 1/4 +.......+99/100 + 1/100

obtinem  (1/2 + 1/2 ) + (2/3 + 1/3) + ( 3/4 + 1/4) +......+(99/100 + 1/100) = 1+1+1+1+.......+1=99