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`.

Answer» Given `u_(1) = 1, u_(2) = 1,u_(n+2) = u_(n+1) + u_(n) n ge 1`
`u_(n) = 1/(sqrt(5))[((1+sqrt(5))/(2))^(n) - ((1-sqrt(5))/(2))^(n) ]`
`:. u_(2) = (1)/(sqrt(5))[((1+sqrt(5))/(2))^(2)-((1-sqrt(5))/(2))^(2)]`
`= (1)/(sqrt(5))((1+sqrt(5))/(2)+(1-sqrt(5))/(2))((1+sqrt(5))/(2)-(1-sqrt(5))/(2))`
`= (1)/(sqrt(5)) (1 xx sqrt(5)) = 1`
Thus m `u_(n)` is true for `n = 2`
Let it be true for `n = k gt 2`" "(as given `u_(2) = 1`). Then
`u_(k) = (1)/(sqrt(5))[((1+sqrt(5))/(2))^(k) - ((1-sqrt(5))/(2))^(k)]`
Consider that ` u_(k+1) = u_(k) + u_(k-1)` [Using `u_(n+2) = u_(n+1) + u_(n)`] ltbr gt `:. u_(k+1) = 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k) - ((1-sqrt(5))/(2))^(k)]`
`+ (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1)-((1-sqrt(5))/(2))^(k-1)]`
`= (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k) + ((1+sqrt(5))/(2))^(k-1)]`
`- [((1-sqrt(5))/(2))^(k) + ((1+sqrt(5))/(2))^(k-1)]`
`= 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1){(1+sqrt(5))/(2) +1}]`
`- [((1-sqrt(5))/(2))^(k-1){(1-sqrt(5))/(2)+1}]`
`= (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1){(6+2sqrt(5))/(4)}-((1-sqrt(5))/(2))^(k-1){(6-2sqrt(5))/(4)}]`
`= 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1)((sqrt(5)+1)/(2))^(2)-((1-sqrt(5))/(2))^(k-1)((sqrt(5)-1)/(2))^(2)]`
`= (1)/(sqrt(5))[((1+sqrt(5))/(2))^(k+1)-((1-sqrt(5))/(2))^(k+1)]`
So, `u_(k+1)` is also true.
Hence, by principle of mathematical induction `u_(n)` is true `AA n gt 1`.


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