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)=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\ a n d\ U_(n+2)=U_(n+1)+U_n for\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\ a n d\ U_(n+2)=U_(n+1)+U_n for\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 for nge1 . Use Mathematical Induction to show that: U_n= 1/sqrt5[((1+sqrt5)/2)^n-((1-sqrt5)/2)^n] for all n ge1 .

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 .

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 .

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 .

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 .

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 U_n=2 cosntheta then U_1U_n-U_(n-1) is equal to