Numerele A1,A2,A3,....,An sunt invers proportionale cu numerele 2,6,12....,n(n+1).Daca A1+A2+A3+....+An=5400 si An-1-An=60,aflati n si An

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am facut niste greseli de redactare...... scuze !!!1 ......voi reveni cu ERATA!!!!!!

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2014-08-24T12:47:33+03:00

2a1 = 6a2 = 12a3 =................=.n(n+1)an                                                   

a2 = a1/3 ; a3 = 1/2.3 ×a1;  a4 = 1/3.4×a1.......... an = 1/n(n+1)

a(n-1) - an = 1/n(n-1) - 1/n(n+1)=(n+1 -n +1)/n(n² -1) =2/ n(n²-1) =60

⇒   1/n(n-1)(n+1) = 60 ⇒   n(n-1)(n+1) = 1/ 3.4.5 ⇒n=3   

a1 + 1/1.3×a1 + 1 /2.3×a1 =5400 ⇒   a 1(1+ 1/3 + 1/6) =a1×3/2 = 5400 ⇒                     a1 = 5400×2/3 = 3600       a3 =3600/6 = 600                              

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2a1=6a2=12a3=.........= n(n+1)an sau, a1=3a2=6a3=.......= n(n+1)/2. a2 =a1/(1+2) a3 = a1/(1+2+3) ....... an= 1/(1+2+3+...+n) a1+a1/(1+2) + a1/(1+2+3)+........... +a1/(1+2+3+.....+n)= 5400 a1[ 1+ 1/(1+2) +1/(1+2+3) +.......+1/(1+2+3+...+n)]=5400 notam S=1 + 1/(1+2) +1/(1+2+3)+..........1/(1+2+3+......n) si observam ca 1+2 =2.3/2 1+2+3 =3.4/2 1+2+3+4=4.5/2 1+2+3+.....+n=n(n+1)/2 deci, S=1+2/2.3 +2/3.4+ 2/4.5 +...2/n(n+1) = 1+2( 1/2.3 +1/3.4 +1/4.5+....+1n(n+1)
S=1+2( 1/2-1/3+1/3-1/4+1/4-1/5+.......+1/(n)- 1/(n+1) =1+2( 1/2-1/(n+1) =2 - 2/(n+1) = =2n/(n+1) 2n.a1/(n+1) =5400 (1) a(n-1) - an =60 2a1/n(n-1) -2a1/n(n+1) =60 a1[1/n(n-1) - 1/n(n+1)]= 30 a1[(n+1-n+1)/n(n-1)(n+1)] a1.2/ n(n-1)(n+1)=30 a1.1/n(n-1)(n+1)=15 (2) (1):(2) = 2n^2.(n-1) = 5400/15 n^2. (n-1)=180 =6^2 x5 si de aici: n-1= 5 n=6 din(2) a1/6.5.7 = 15 a1= 3150 an=a6= 3150/(1+2+...6)= 3150:21= 150