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If (9+4sqrt(5))^(n)=p+beta, where n and p are positive integers and beta is a positive proper fraction, prrove that (1-beta)(p+beta)=1 and p is an odd integer. |
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Answer» <P> Solution :We have,`(9+4sqrt(5))^(n)=p+beta,p` is a +ve integer, `oltbetalt1` LET `(9-4sqrt(5))^(n)lt1` ALSO `(9-4sqrt(4))^(n)gt0` `therefore0ltflt1` now, `p+beta+f=(9+4sqrt(5))^(n)+(9-4sqrt(5))^(n)=2[.^(n)C_(0)9^(n)+.^(n)C_(2)9^(n-2)80+.^(n)C_(4)9^(n-4)(80)^(2)+ . . . .]` `impliesp+beta+f`=even integer `impliesbeta+f=`even integer -p=even integer-integer=an integer. since `0 lt beta lt 1 and 0 lt f lt 1` `implies 0 lt beta+ f lt 2, ` but `beta+f=` an integer `impliesbeta+f=1impliesf=1-beta` Now `p+beta+f=`ven integer `impliesp`=even integer-1= odd integer annd `(p+beta)f=(9+4sqrt(5))^(n)(9-4sqrt(5))^(n)` `=(81-80)^(n)=1` `implies(p+beta)(1-beta)=1`. |
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