Question

Folosind relația p supra n•(n+p)=1 supra n - 1 supra n+p , calculati suma 3 supra 1×4 + 3 supra 4×7 + 3 supra 7×10 + ... + 3 supra 55×58.

Answer 1

p/[n(n+p)] = 1/n - 1/(n+p)
S = 3/(1·4 )+ 3/(4·7) + 3/(7·10) + .......+ 3/(55/58)
S = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 +.........+1/55 - 1/58 = 1/1 - 1/58 = 57/58

Answer 2

Observi ca toti termenii din acea suma sunt de forma din acea relatie unde n ia valori de la 1 la 55 si p este intotdeauna 3
$\frac{3}{1*4}=\frac{3}{1(1+3)}=\frac{1}{1}-\frac{1}{4}$
$\frac{3}{4*7}=\frac{3}{4(4+3)}=\frac{1}{4}-\frac{1}{7}$
$\frac{3}{7*10}=\frac{3}{7(7+3)}=\frac{1}{7}-\frac{1}{10}$

$\frac{3}{52*55}=\frac{3}{52(52+3)}=\frac{1}{52}-\frac{1}{55}$
$\frac{3}{55*58}=\frac{3}{55(55+3)}=\frac{1}{55}-\frac{1}{58}$
Notam acea suma cu S. Ne apucam sa adunam toate relatiile de mai sus si in stanga egalitatii obtinem suma ceruta
$S=\frac{3}{1*4}+\frac{3}{4*7}+\frac{3}{7*10}+..._\frac{3}{52*55}+\frac{3}{55*58}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{52}-\frac{1}{55}+<span>\frac{1}{55}-\frac{1}{58}=1-\frac{1}{58}=\frac{58-1}{58}\frac{57}{58}</span>$