化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求:化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求2^99+2^98+2^97+······+2+1的末尾数字末尾数字末尾数字末尾数字末尾数字末尾数

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化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求:化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求2^99+2^98+2^97+······+2+1的末尾数字末尾数字末尾数字末尾数字末尾数字末尾数
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化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求:化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求2^99+2^98+2^97+······+2+1的末尾数字末尾数字末尾数字末尾数字末尾数字末尾数
化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求:
化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求2^99+2^98+2^97+······+2+1的末尾数字
末尾数字末尾数字末尾数字末尾数字末尾数字末尾数字!

化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求:化简(x+1)(x^99+x^98+······+x+1),我已用规律求出x^100-1,求2^99+2^98+2^97+······+2+1的末尾数字末尾数字末尾数字末尾数字末尾数字末尾数
(1)(a-1)(a+1)=a^2-1; (2)(a-1)(a^2+a+1)= a^3-1;
(3)(a-1)(a^3+a^2+a+1)= a^4-1;……
由此我们可以得到:
(a-1)(a^99+a^98+a^97+…+a+1)=a^100-1
请你利用上面的结论,完成下面两题的计算:
(1)2^99+2^98+2^97+…+2+1;
=(2-1)(2^99+2^98+.+2+1)
=2^100-1
末尾数字 2 4 8 6 每四次方循环一次
100/4=25
末尾为 6-1 =5

先计算下列各式的值:
(1)(x-1)(x+1)=x^2-1
(2)(x-1)(x^2+x+1)=x^3-1
(3)(x-1)(x^3+x^2+x+1)=x^4-1
......
由此可以得出:(x-1)(x^99+x^98+x^97+...+x+1)=x^100-1
∴x^99+x^98+x^97+...+x+1=(x^100-1)/(x...

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先计算下列各式的值:
(1)(x-1)(x+1)=x^2-1
(2)(x-1)(x^2+x+1)=x^3-1
(3)(x-1)(x^3+x^2+x+1)=x^4-1
......
由此可以得出:(x-1)(x^99+x^98+x^97+...+x+1)=x^100-1
∴x^99+x^98+x^97+...+x+1=(x^100-1)/(x-1)
∴(1)2^99+2^98+2^97+...+2+1
=(2^100-1)÷(2-1)=2^100-1
(2)(-2)^50+(-2)^49+(-2)^48+...+(-2)+1
=〔(-2)^51-1〕÷(-2-1)
=(2^51+1)/3

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