Question

0,6,-6,18,-30,66…;第n个数是多少?

Answer 1

a2-a1=6
a3-a2=-12
a4-a3=24
a5-a4=-48
a6-a5=96
.
所以an - an-1 = -2*(an-1 - an-2) = ...= (-2)^(n-2)(a2 - a1) = 6*(-2)^(n-2)
即 an=an-1 + 6*(-2)^(n-2) = an-2 + 6*(-2)^(n-3) + 6*(-2)^(n-2) = ...= a1 + 6*(-2)^0 + 6*(-2)^1 + ...+ 6*(-2)^(n-2) = 6*(1-(-2)^(n-1))/(1-(-2)) = 2*( 1 - (-2)^(n-1) )
最后验算一下有没算错：
a1 = 2 * (1 - (-2)^(1-1) ) = 0 对的
a2 = 2 * (1 - (-2)^(2-1) ) = 2 * (1 - (-2)) = 6 对的
.
a6 = 2 * (1 - (-2)^(6-1) ) = 2 * (1 - (-32)) = 66 对的
.
所以第n项应该就是
an = 2 * ( 1 - (-2)^(n-1) )