Question

a(n+1)-an=n+2^n 且a1=1,求an

Answer 1

a(n+1)-an=n+2^n

an-an-1=n-1+2^(n-1)
an-1-an-2=n-2+2^(n-2)
an-2-an-3=n-3+2^(n-3)
:
a2-a1=1+2^1
等式两边相加得
an-a1=(1+2+3+……+n-1)+(2^1+2^2+……+2^(n-1)
an=n(n-1)/2+2^n -2+1=2^n+n(n-1)/2-1

Answer 2

an-a(n-1) = n-1 + 2^(n-1)
...
a2-a1 = 1 + 2
a1 = 1

an = 1 + (1+2+3+...+n-1) + (2 + 4 + ... + 2^(n-1) )
= 1 + n(n-1) / 2 + 2^n - 2
= 2^n + n(n-1)/2 -1

Answer 3

an=[an-a(n-1)]+[a(n-1)-a(n-2)]+```+(a2-a1)+a1
=[(n-1)+2^(n-1)]+[(n-2)+2^(n-2]+```+(1+2)+1
=[2^(n+1)+n^2-n-2]\2

Answer 4

做差法。
就是
a2-a1=1+2^1
a3-a2=2+2^2
an-a(n-1)=(n-1)+2^(n-1)
以此类推，等号两边同时相加，
得an=2^n + n(n-1)/2 -1 .