b1=2,b(n+1)=(3bn+4)/(2bn+3),求bn.

Question

b1=2,b(n+1)=(3bn+4)/(2bn+3),求bn.

Answer 1

b1=2,b(n+1)=(3bn+4)/(2bn+3),
b(n+1)+√2=(3bn+4)/(2bn+3)+√2
=(3bn+4+2√2bn+3√2)/(2bn+3)
=(3+2√2)(bn+√2)/(2bn+3)
b(n+1)-√2=(bn+4-2√2bn-3√2)/(2bn+3)
=(3-2√2)(bn-√2)/(2bn+3)
所以[b(n+1)+√2]/[b(n+1)-√2]=[(3+2√2)/(3-2√2)][(bn+√2)/(bn-√2)]
=(3+2√2)²[(bn+√2)/(bn-√2)]
所以{(bn+√2)/(bn-√2)}是公比为(3+2√2)²的等比数列
首项=(b1+√2)/(b1-√2)=(3+2√2)
于是(bn+√2)/(bn-√2)=(3+2√2)*(3+2√2)^2(n-1)=(3+2√2)^(2n-1)
所以an=√2*[(3+2√2)^(2n-1)+1]/[(3+2√2)^(2n-1)-1]