Question Number 101473 by 175 last updated on 02/Jul/20

Answered by 1549442205 last updated on 03/Jul/20

  We prove by induction method that  ∃m∈N such that   A_n =(2+(√2))^n =(m_n +(√(m_n ^2 −2^n )))and q_n =(√((m_n ^2 −2^n )/2))∈N  A_n =m_n +q_n (√2)  +For n=1 we have 2+(√2)=2+(√(2^2 −2))  and (√((2^2 −2)/2))=1∈N⇒State true  +Suppose State was true for ∀n≤k  means that we had A_k =(2+(√2))^k =(m_k +(√(m_k ^2 −2^k )))  and q_k =(√((m_k ^2 −2^k )/2))∈N,A_k =m_k +q_k (√2)  +Then A_(k+1) =(2+(√2))^(k+1) =(2+(√2))^k (2+(√2))  =A_k (2+(√2))=(m_k +(√(m_k ^2 −2^k )))(2+(√2))  =2m_k +(√(2m_k ^2 −2^(k+1) ))+(√2)m_k +2(√(m_k ^2 −2^k ))  =(2m_k +2(√((m_k ^2 −2^k )/2)))+(√2)(m_k +(√(2m_k ^2 −2^(k+1) )))  =2m_k +2q_k +(m_k +2q_k )(√2)=2(m_k +q_k )+(√(2(m_k +2q_k )^2 ))  =2(m_k +q_k )+(√(2m_k ^2 +8m_k q_k +8q_k ^2 ))  =2(m_k +q_k )+(√(2m_k ^2 +8m_k q_k +4q_k ^2 +4.((m_k ^2 −2^k )/2)))  =2(m_k +q_k )+(√((4m_k ^2 +8m_k q_k +4q_k ^2 )−2^(k+1) ))  =2(m_k +q_k )+(√(4(m_k +q_k )^2 −2^(k+1) .))  Putting m_(k+1) =2(m_k +q_k ),q_(k+1) =m_k +2q_k    then A_(k+1) =m_(k+1) +(√(m_(k+1) ^2 −2^(k+1) ))and   q_(k+1=) (√(((m_(k+1) ^2 −2^(k+1) )/2) )) ∈N which shows  that State is true for n=k+1  By indution principle State is true   ∀n∈N .Thus,there is always exists a  natural number m so that  (2+(√2))^n =(m+(√(m^2 −2^n )) ∀n∈N (q.e.d)

Commented by175 last updated on 03/Jul/20

thank, please answer the quation 2

Commented by175 last updated on 03/Jul/20

khd

Commented by1549442205 last updated on 04/Jul/20

You are welcome sir.