Question

用分解因式法求(-2）^1999+(-2)^1998+（-2）^1997+……+（-2）^3+(-2)^2+(-2)+1的值

Answer 1

设s=(-2）^1999+(-2)^1998+（-2）^1997+……+（-2）^3+(-2)^2+(-2)+1

-2s=(-2)^2000+(-2）^1999+(-2)^1998+（-2）^1997+……+（-2）^3+(-2)^2+(-2)
-2s-s=(-2）^2000+1
s=-1/3*2^2000-1/3

Answer 2

(-2)^(2n+1) + (-2)^(2n)
=(-2)^(2n)*(-2)+ (-2)^(2n)
=(-2)^(2n)(-2+1)
=-(-2)^(2n)
=-2^2n
所以原式=Σ(-2)^(2n+1) + (-2)^(2n), n = 0 .... 999
=-Σ2^2n, n = 0 .... 999
考察序列
Σ2^m, m = 0 , 2n+1
=Σ2^2n +Σ2^(2n+1)
而Σ2^(2n+1)/Σ2^2n = 2
所以Σ2^2n/Σ2^m = 1/3
Σ2^m
= 2^mΣ(1+1/2+1/4+....+1/2^m)
=2^m(2-1/2^m)
=2^(m+1)-1
Σ2^2n = 1/3Σ2^m
=(2^(m+1)-1)/3
以上式中n=999,m=2n+1=1999
=(2^2000-1)/3
所以原式= - Σ2^2n
= - (2^2000-1)/3

Answer 3

(- 2)^1999 + (- 2)^1998 + (- 2)^1997 + ... + (- 2)^3 + (- 2)^2 + (- 2) + 1
= (- 2)^(1999 + 1998 + 1997 + ... + 1) + 1
= (- 2)^(1999(1999 + 1)/2) + 1
= (- 2)^(1999 * 1000) + 1
= 2^1999000 + 1

Answer 4

就是-2的（1999+1998+1997+......+2+1)次方再加1
也就是-2的1999000次方加1