Question

对任意x∈R,|2-x|+|3+x|≥a²-4a恒成立,则a的取值范围

Answer 1

|2-x|+|3+x|≥|2-x+3+x|=5
所以只要a²-4a≤5
a²-4a-5=(a+1)(a-5)≤0
-1≤a≤5

Answer 2

Sn=(16-2^1)+(16-2^2)+……+(16-2^n)
=16+16+……+16-(2^1+2^2+……+2^n)
=16n-2*(1-2^n)/(1-2)
=16n+2-2^(n+1)