Answer to Question #33699 in Real Analysis for soniya l

Question

let a1=a2=5 and an+1=an +6an-1 ,n>=2.show by induction that an=3n-(-2)n if n>=1.

Answer 1

a1=5, a2=5, a3=a2+6a1=35n=2 then a_2=3^(2) - (-2)^2=9-4=5n=3 then a_3= 3^(3) - ( -2)^3=27-(-8)=35 So we have the base of induction. Now let's suppose that for all n<=k a_n=3^(n)-(-2)^(n) We must prove that a_(k+1)=3^(k+1)-(-2)^(k+1) we know that a_(k+1)=a_(k)+6*a_(k-1) but for k and k-1 indeces we have formula held so a_k=3^(k)-(-2)^(k) , a_(k-1)=3^(k-1)-(-2)^(k-1) Now a_(k+1)=a_(k)+6*a_(k-1)=3^(k)-(-2)^(k) +6*( 3^(k-1)-(-2)^(k-1) )= 3^(k)-(-2)^(k) +6* 3^(k-1)-6*(-2)^(k-1) = = 3^(k)-(-2)^(k) +2*3* 3^(k-1)-(-3)*(-2)*(-2)^(k-1)=3^(k)-(-2)^(k) +2*3^(k)+3*(-2)^(k)=3*3^(k) +2*(-2)^(k)=3^(k+1) - (-2)*(-2)^(k)=3^(k+1)-(-2)^(k+1)Thus, by induction we prove formula for a_n