Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 101247 by 175 last updated on 01/Jul/20

Answered by 1549442205 last updated on 06/Jul/20

Note :the integral part of a real number x be the grearest integer which don′t exceed x and is denoted by symbol [x] Put A_n = (2+(√2))^n +(2−(√2))^n .Then we prove that A_n is even ∀n∈N .Indeed, i)For n=0 ⇒ A_0 =2.For n=1⇒A_1 =4 A_n =[(2+(√2))^n +(2−(√2))^n ]= [(2+(√2))^(n−1) +(2−(√2))^(n−1) ][(2+(√2))+(2−(√2))] −(2−(√2))(2+(√2))^(n−1) −(2+(√2))(2−(√2))^(n−1) =4.A_(n−1) −2(2+(√2))^(n−2) −2(2−(√2))^(n−2) =4.A_(n−1) −2A_(n−2) which shows that A_n is an even number ∀n∈N Now we see that 0<m=(2−(√2))^n <1 due to 2−(√2)<1.It follows that (2+(√2))^n =A_n −(2−(√2))^n =2M−m ⇒[(2+(√2))^n ]=2M−1 which means that [2+(√2))^n ]is odd(q.e.d)

Note:theintegralpartofarealnumberxbethegrearestintegerwhichdon′texceedxandisdenotedbysymbol[x]PutAn=(2+2)n+(2−2)n.ThenweprovethatAniseven∀n∈N.Indeed,i)Forn=0⇒A0=2.Forn=1⇒A1=4An=[(2+2)n+(2−2)n]=[(2+2)n−1+(2−2)n−1][(2+2)+(2−2)]−(2−2)(2+2)n−1−(2+2)(2−2)n−1=4.An−1−2(2+2)n−2−2(2−2)n−2=4.An−1−2An−2whichshowsthatAnisanevennumber∀n∈NNowweseethat0<m=(2−2)n<1dueto2−2<1.Itfollowsthat(2+2)n=An−(2−2)n=2M−m⇒[(2+2)n]=2M−1whichmeansthat[2+2)n]isodd(q.e.d)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com