Prove without mathematical induction that the expression (1 + (√2))^(2n) + (1 − (√2))^(2n) is even for every natural number n.


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Question Number 58409 by Tawa1 last updated on 22/Apr/19

$Prove without mathematical induction that the$ $expression {(1 + \sqrt{2})}^{2 n} + {(1 - \sqrt{2})}^{2 n} is even for every$ $natural number n .$
Commented by maxmathsup by imad last updated on 22/Apr/19
$let A_n =(1+(√2))^(2n) +(1−(√2))^(2n) ⇒A_n =Σ_(k=0) ^(2n) C_(2n) ^k ((√2))^k +Σ_(k=0) ^(2n) C_(2n) ^k (−(√2))^k =Σ_(k=0) ^(2n) C_(2n) ^k { ((√2))^k +(−(√2))^k } =Σ_(k =2p) (...) +Σ_(k=2p+1) (...) =Σ_(p=0) ^n C_(2n) ^(2p) 2^(p+1) +0 =2 {Σ_(p=0) ^n C_(2n) ^(2p) 2^p } ⇒A_n is even .$
$l e t A_{n} = {(1 + \sqrt{2})}^{2 n} + {(1 - \sqrt{2})}^{2 n} \Rightarrow A_{n} = \sum_{k = 0}^{2 n} C_{2 n}^{k} {(\sqrt{2})}^{k} + \sum_{k = 0}^{2 n} C_{2 n}^{k} {(- \sqrt{2})}^{k}$ $= \sum_{k = 0}^{2 n} C_{2 n}^{k} {{(\sqrt{2})}^{k} + {(- \sqrt{2})}^{k}} = \sum_{k = 2 p} (. . .) + \sum_{k = 2 p + 1} (. . .)$ $= \sum_{p = 0}^{n} C_{2 n}^{2 p} 2^{p + 1} + 0 = 2 {\sum_{p = 0}^{n} C_{2 n}^{2 p} 2^{p}} \Rightarrow A_{n} i s e v e n .$
Commented by Tawa1 last updated on 23/Apr/19

$God bless you sir, i will check if i can understand the solution$
Commented by maxmathsup by imad last updated on 23/Apr/19

$y o u a r e w e l c o m e s i r, h e r e i h a v e u s e d t h e b i n o m e f o r m u l a e .$
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