calculam intai suma fractiilor din paranteza, unde observam ca numitorii sunt produse a cate doua numere impare consecutive de forma:
1/(2k+1)(2k+3)
vom rescrie fiecare termen al cestei sume astfel:
1/1x3 devine 1/2x(3-1)/1x3; obervam ca aceasta scriere nu modifica valoarea termenului initial, doar ajuta ulterior la calculul sumei
la fel: 1/3x5 devine 1/2x(5-3)/3x5,
1/5x7 devine 1/2x(7-5)/5x7.
1/7x9 devine 1/2x(9-7)/7x9
s.a.m.d......................rescriem penultimul si ultimul termen al sumei:
1/59x61 devine1/2x(61-59)/59x61
1/61x63 devine 1/2(63-61)/61x63
inlocuind termenii sumei cu noua formulare si observand ca fiecare termen are factorul 1/2 comun, obtinem:
1/2[(3-1)/1x3+(5-3)/3x5+(7-5)/5x7+(9-7)/7x9+.....................+(61-59)59x61+ (63-61)61x63]=
1/2[3/(1x3)-1/(1x3)+5(3x5)-3/(3x5)+7/(7x5)-5/(7x5)+9/(7x9)-7/(7x9)+
+..............61/(59x61)-59/(59x61)+63/(61x63)-61/(61x63)]
observam ca la fiecare termen se pot face simplificari astfel:
3/(1x3) devine 1
1/(1x3) ramane 1/3
5/(3x5) devine 1/3
3/(3x5) devine 1/5
7/(7x5) devine 1/5
..........................
63/(61x63) devine 1/61
61/(61x63) devine 1/63
observam ca termenii sumei se repeta o data cu + o data cu -,
deci toti acesti termeni din suma se vor reduce la zero;
raman doar doi termeni: 1-1/63
Deci suma din paranteza devine:
1/2[(1-1/63)]=1/2(63/63-1/63)=1/2x62/63;
simplificam cu 2 si obtinem suma din paranteza 31/63
inlocuim in relatia initiala si obtinem:
X x 31/63 = 2-64/63
(31xX)/63 = 2x63/63-64/63
(31xX)/63 =126/63-64/63
(31xX)/63= 62/63
31xX=62
X=62:31
X=2
