Let `U_1=1,\ U_2=1\ a n d\ U_(n+2)=U_(n+1)+U_nfor\ ngeq1.` use mathematical induction to show that: `U_n=1/(sqrt(5)){((1+sqrt(5))/2)^n-\ ((1-sqrt(5))/2)^n}\ for\ a l l\ ngeq1.`

Let U_(1)=1,U_(2)=1 and U_(n+2)=U_(n+1)+U_(n)f or n>=1 use mathematical induction to show that: U_(n)=(1)/(sqrt(5)){((1+sqrt(5))/(2))^(n)-((1-sqrt(5))/(2))^(n)}f or all n>=1

Let U_(1)=1,U_(2)=1 and U_(n+2)=U_(n+1)+U_(n) for n>=1. Use mathematical induction to such that: :U_(n)=(1)/(sqrt(5)){((1+sqrt(5))/(2))^(n)-((1-sqrt(5))/(2))^(n)} for all n>1

Let u_(1)=1,u_2=2,u_(3)=(7)/(2)and u_(n+3)=3u_(n+2)-((3)/(2))u_(n+1)-u_(n) . Use the principle of mathematical induction to show that u_(n)=(1)/(3)[2^(n)+((1+sqrt(3))/(2))^n+((1-sqrt(3))/(2))^n]forall n ge 1 .

If a_(1)=1, a_(2)=5 and a_(n+2)=5a^(n+1)-6a_(n), n ge 1 . Show by using mathematical induction that a_(n)=3^(n)-2^(n)

Use mathematical induction to show that 1+3+5+…+ (2n-1) = n^(2) is true for a numbers n.

Let u_(n)=(1)/(sqrt((5)))[((1+sqrt(5))/(2))^(n)-((1-sqrt(5))/(2))^(n)] (0=0,1,2,3,……) , prove that u_(n+1)=u_(n)+u_(n-1)(n ge 1) .

If n is a positive interger,them (sqrt(5)+1)^(2n+1)-(sqrt(5)-1)^(2n-1) is

If a_1=1\ a n d\ a_(n+1)=(4+3a_n)/(3+2a_n),\ ngeq1\ a n d\ if\ (lim)_(n->oo)a_n=n then the value of a_n is sqrt(2) b. -sqrt(2) c. 2\ d. none of these

If U= [((1)/(sqrt2),(-1)/(sqrt2)),((1)/(sqrt2),(1)/(sqrt2))] , then U^(-1) is

int (du) / (u sqrt (u ^ (2) -1))