Question

已知数列an满足a1=1,an+1=2an+3的n次方,求an

Answer 1

利用待定系数法，令
a(n+1)+A*3^(n+1)=2*(an+A*3^n)
a(n+1)=2an-A*3^n
所以A=-1
即a(n+1)-3^(n+1)=2(an-3^n)
因为a1-3^1=1-3=-2
所以an-3^n=(-2)*2^(n-1)=-2^n
an=3^n-2^n

Answer 2

见图

Answer 3

an+1=2an+2的n次方-1 两边同除以 2^(n+2) an+1/ 2^(n+2) = (2an+2^(n-1))/2^(n+2) = an/2^(n+1) + 1/8 an+1/ 2^(n+2) - an/2^(n+1) = 1/8 所以an/2^(n+1) 为等差数列，即an-1/2^n 为等差数列 设bn = an/2^(n+1) b1 = a1/2^2 =1/4 bn = b1+1/8 *(n-1) = 1/4 +1/8 n - 1/8 = 1/8 n + 1/8 an = 2^(n+1) * bn = 2^(n+1) * (1/8 n + 1/8) =n 2^(n-2) + 2^(n-2)