Question

(a+1)(a2+1)(a4+1).......(a1024+1)

初一数学题

Answer 1

(a+1)(a2+1)(a4+1).......(a1024+1)
a不等于1时
=[(a-1)(a+1)(a2+1)(a4+1).......(a1024+1)]/(a-1)
=[（a^2-1）(a2+1)(a4+1).......(a1024+1)]/(a-1)
=[(a4-1)(a4+1).......(a1024+1)]/(a-1)
……
=[a^2048-1)]/(a-1)
a=1时(a+1)(a2+1)(a4+1).......(a1024+1)=2^1024

Answer 2

当a=1时，原式=2*2*2*......*2=2^11
当a不等于1 ，在原式中，乘以（a-1）,在除以（a-1）。运用平方差公式合并，有
(a+1)(a^2+1)(a^4+1).......(a^1024+1)（a-1）/（a-1）
=[(a-1)(a+1)](a^2+1)(a^4+1).......(a^1024+1)/(a-1)
=(a^2-1)(a^2+1)(a^4+1).......(a^1024+1)/(a-1)
=......=(a^2048-1)/(a-1)