Question Number 101474 by 175 last updated on 02/Jul/20
Answered by 1549442205 last updated on 07/Jul/20
![(see the solution of the question 101247) we known that [(2+(√2))^n ]=(2+(√2))^n +(2−(√2))^n −1 is an odd number.Putting A_n =(2+(√2))^n +(2−(√2))^n then a_n =A_n −1.Using induction method +For n=2 ⇒a_2 =[(2+(√2))^2 ]=[6+4(√2)]=11 =2.3+1+(3+1)=2a_1 +1+(a_1 +1) ⇒Equality true +Suppose Equality is true ∀n≤k which means we had a_k =2a_(k−1) +1+𝚺_(r=1) ^(k−1) (a_r +1) +Then a_(k+1) =[(2+(√2))^(k+1) ]=A_(k+1) −1 =(2+(√2))^(k+1) +(2−(√2))^(k+1) −1=[(2+(√2))^k +(2−(√2))^k ][(2+(√2) )+(2−(√2))] −2(2+(√2))^(k−1) −2(2−(√2))^(k−1) −1 =4A_k −2A_(k−1) −1=4(A_k −1)−2(A_(k−1) −1)+1 =4a_k −2a_(k−1) +1=2a_k +1+a_k +(a_k −2a_(k−1) ) =2a_k +1+a_k +{[2a_(k−1) +1+𝚺_(r=1) ^(k−1) (a_r +1)]−2a_(k−1) } =2a_k +1+(a_k +1+𝚺_(r=1) ^(k−1) (a_r +1)) =2a_k +1+𝚺_(r=1) ^(k) (a_r +1).This shows that Equality is also true for n=k+1.Hence, by induction Equality is true ∀n∈N,n>1 (q.e.d)](Q101584.png)