若(2x+√5)^5=a0+a1x+a2x^2+a3x^3+a4x^4+a5x^5,则(a0+a2+a4)^2-(a1+a3+a5)^2=

Question

若(2x+√5)^5=a0+a1x+a2x^2+a3x^3+a4x^4+a5x^5,则(a0+a2+a4)^2-(a1+a3+a5) ^2=

Answer 1

令x=1,a0+a1+a2+a3+a4+a5=(2+√5)⁵ ①
令x=-1,a0-a1+a2-a3+a4-a5=(-2+√5)⁵ ②
①+②得,2a0+2a2+2a4=(2+√5)⁵ +(-2+√5)⁵
①-②得,2a1+2a3+2a5=(2+√5)⁵ -(-2+√5)⁵
(a0+a2+a4)²-(a1+a3+a5)²
=[(2+√5)⁵+(-2+√5)⁵]²-[(2+√5)⁵-(-2+√5)⁵]²
=[(2+√5)⁵+(-2+√5)⁵+(2+√5)⁵-(-2+√5)⁵][(2+√5)⁵+(-2+√5)⁵-(2+√5)⁵+(-2+√5)⁵]
=[2×(2+√5)⁵][2×(-2+√5)⁵]
=4[(2+√5)(-2+√5)]⁵
=4×(5-1)⁵
=4