求-12+36-108+.+4x3的100次方的值

设A=-12+36-108+.+4x3的100次方
=4×(-3)^1+4×(-3)^2+4×(-3)^3+……+4×(-3)^100
=4×[(-3)^1+(-3)^2+(-3)^3+……+(-3)^100]
令B=-3A
则B=4×[(-3)^2+(-3)^3+(-3)^4+……+(-3)^101]
令A-B=A+3A=4A
=4×[(-3)^1-(-3)^101]
即4A=4×[(-3)^1-(-3)^101]
所以,A=(-3)^1-(-3)^101
=3^101-3