Question

数列的通项问题

这个1 -1 -5 -13 -29······数列的通项公式是？

Answer 1

a1=1,a2=-1,a3=-5,a4=-13,a5=-29

a2-a1=-2=-(2)^1
a3-a2=-4=-(2)^2
a4-a3=-8=-(2)^3
a5-a4=-16=-(2)^4
....
an-a(n-1)=-(2)^(n-1)

所以an=a1+(a2-a1)+(a3-a2)+...+(an-a(n-1))
an=1-(2+4+8+...+2^(n-1))
an=-2^n+3

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Answer 2

用累差法可以求

a1 = 1
a2 - a1 = -2
a3 - a2 = -4
a4 - a3 = -8
...
a(n) - a(n-1) = -2^(n-1)

然后左边相加得：
a(n) = 1 -2 -4 -8 -... - 2^(n-1)
= -2^n+3

Answer 3

an+1 =2an -3
an+1-3=2(an-3)
(an+1-3)/(an-3)=2
故an-3=(a1-3)2^n
=-2^(n+1)
故an=-2^(n+1)+3,n>=0

Answer 4

a[n+1]=a[n]-2^(n-1)
方括号{}内为下标