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Un-1-2-n-show-that-we-have-p-n-N-U-n-p-n-p-n-1-




Question Number 102474 by pticantor last updated on 09/Jul/20
Un=(1+(√2))^n   show that we have p_n ∈N /  U_n =(√p_n )+(√(p_n +1))
Un=(1+2)nshowthatwehavepnN/Un=pn+pn+1
Answered by ~blr237~ last updated on 09/Jul/20
 observe that (1/U_n )=(−1)^n (1−(√2))^n   and that there exist   a_n  ,b_n ∈N  such as   U_n =a_n +b_n (√2)    and  (1/U_n )=(−1)^n (a_n −b_n (√2))  and  U_n ^2 =U_(2n)  .  So  (U_n ^ +(1/U_n ))^2 = 2+U_(2n) +(1/U_(2n) )=2 +2a_(2n)    (U_n −(1/U_n ))^2 =−2+U_(2n) +(1/U_(2n) ) =−2+2a_(2n)   we have  a_(n+1) +b_(n+1) (√2) =(1+(√2))(a_n +b_n (√2) ) lead to  a_(n+1) =a_n +2b_n  . So  such as a_1 =1 (odd ) , a_n  is always odd   So there exist  c_n  ∈N  / a_n =2c_n +1  Then    (U_n +(1/U_n ))^2 =2+2(2c_(2n) +1)=4(c_(2n) +1)   (U_n −(1/U_n ))^2 =−2+2(2c_(2n) +1)=4c_(2n)    By adding the two square root   U_n =(√c_(2n) ) +(√(c_(2n) +1))    take  p_n =c_(2n)
observethat1Un=(1)n(12)nandthatthereexistan,bnNsuchasUn=an+bn2and1Un=(1)n(anbn2)andUn2=U2n.So(Un+1Un)2=2+U2n+1U2n=2+2a2n(Un1Un)2=2+U2n+1U2n=2+2a2nwehavean+1+bn+12=(1+2)(an+bn2)leadtoan+1=an+2bn.Sosuchasa1=1(odd),anisalwaysoddSothereexistcnN/an=2cn+1Then(Un+1Un)2=2+2(2c2n+1)=4(c2n+1)(Un1Un)2=2+2(2c2n+1)=4c2nByaddingthetwosquarerootUn=c2n+c2n+1takepn=c2n

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