Question

E(x)=[1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4)] • (x+4)

Answer 1

aducem la acelasi numitor:

= {[(x+3)(x+4) +(x+1)(x+4 + (x+1)(x+2)] / (x+1)(x+2)(x+3)(x+4)} × (x+4)

= (x² +4x +3x +12 +x² +4x +x +4 +x² +2x +x +2) / (x+1)(x+2)(x+3)

= (3x² +15x +18) / (x+1)(x+2)(x+3) = 3(x² +5x + 6) / (x+1)(x+2)(x+3)

= 3( x² + 2x +3x +6) / (x+1)(x+2)(x+3) = 3[x(x+2) + 3(x+2)] / (x+1)(x+2)(x+3)

= 3(x+2)(x+3) / (x+1)(x+2)(x+3 = 3 / (x+1)