An+2=2A(n+1)+15An两边加kA(n+1)得
工A(n+2)+kA(n+1)=(2+k)A(n+1)+15An
=>1/(2+k)=k/15=>k^2+k-15=0=>k=-5或k=3
①k=-5时得(A(n+2)-5A(n+1))/( A(n+1)-5An)=-3验证A1,A2正确,所以λ=-5,公比为-3
②k=3时得(A(n+2)+3A(n+1))/ (A(n+1)+3An)=5验证A1,A2正确,所以λ=3,公比为5
An+2=2A(n+1)+15An两边加kA(n+1)得
工A(n+2)+kA(n+1)=(2+k)A(n+1)+15An
=>1/(2+k)=k/15=>k^2+k-15=0=>k=-5或k=3
①k=-5时得(A(n+2)-5A(n+1))/( A(n+1)-5An)=-3验证A1,A2正确,所以λ=-5,公比为-3
②k=3时得(A(n+2)+3A(n+1))/ (A(n+1)+3An)=5验证A1,A2正确,所以λ=3,公比为5