解答:
1+x+x^2+x^3+…+x^15
=(1+x)+(x^2+x^3)+(x^4+x^5)…+(x^14+x^15)
=(1+x)+x^2(1+x)﹢x^4(1+x)﹢…﹢x^14(1+x)
=(1+x)(x^2+x^4+…+x^14) (以下同理)
=(1+x)(1+x²)(1+x^4+x^8+x^12))
=(1+x)(1+x²)(1+x^4)(1+x^8)
令S=1+x+x^2+x^3+```+x^15,那么:
x*S=x++x^2+x^3+```+x^15+x^16
则有:x*S-S=(x-1)S=x^16 -1
所以:S=1+x+x^2+x^3+```+x^15
=(x^16 -1)/(x-1)
=(x+1)(x²+1)(x^4 +1)(x^8 +1)
原式=(1﹢x)﹢x^2(1﹢x)﹢x^4(1﹢x)﹢…﹢x^14(1﹢x)
=(1﹢x)(x^2﹢x^4﹢…﹢x^14)
=(1﹢x)(1﹢x²)(1﹢x^4﹢x^8﹢x^12)
=(1﹢x)(1﹢x²)(1﹢x^4)(1﹢x^8)
等比数列求和=1*(1-x^16)/(1-x)