在数列an中a1=1an+1=2an+2的n次求a10

Question

在数列an中 a1=1 an+1=2an+2的n次 求a10

Answer 1

a(n+1)=2an+2^n
a(n+1)/2^(n+1)-an/2^n= 1/2
{an/2^n}是等差数列,d=1/2
an/2^n - a1/2^1 = (n-1)/2
an/2^n = n/2
an=n.2^(n-1)
a10=10.2^9 = 5120