Question

观察下列各式(x-1)(x²+x+1)=x²-1，(x-1)(x³+x²+1)=x四次方-1

根据以上算式，求出1+2+2²+·········+X六十二次方+2六十三次方的结果是

Answer 1

题目出的不对吧？
题干应该是(x-1)(x²+x+1)=x3-1，(x-1)(x³+x²+x+1)=x4-1,
所以问题中X=2
（2-1）（1+2+2²+·········+X六十二次方+2六十三次方）=2六十四次方-1

Answer 2

xⁿ - 1 = (x - 1)(xⁿ-¹ + xⁿ-² + ... + x + 1)

追问

根据以上算式，求出1+2+2²+·········+X六十二次方+2六十三次方的结果是

追答

1 + 2 + ... + 2ⁿ = 2ⁿ+¹ - 1
1 + 2 + 2² + ... + 2^63 = 2^64 - 1
1 + x + x² + ... + x^63 = (x^64 - 1) / (x - 1)

追问

``````````````````````````,注意例题是乘，问题是加

追答

(2 - 1)(1 + 2 + 2² + ... + 2^63) = 2^64 - 1
(x - 1)(1 + x + x² + ... + x^63) = x^64 - 1