Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

$$\mathrm{Prove}\:\mathrm{without}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\:\:\:\left(\mathrm{1}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:+\:\left(\mathrm{1}\:−\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:\:\mathrm{is}\:\mathrm{even}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\:\mathrm{n}. \\ $$

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 .

$${let}\:{A}_{{n}} =\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}} \:+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}} \:\:\Rightarrow{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left(\sqrt{\mathrm{2}}\right)^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left(−\sqrt{\mathrm{2}}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left\{\:\left(\sqrt{\mathrm{2}}\right)^{{k}} \:+\left(−\sqrt{\mathrm{2}}\right)^{{k}} \right\}\:=\sum_{{k}\:=\mathrm{2}{p}} \:\:\left(...\right)\:+\sum_{{k}=\mathrm{2}{p}+\mathrm{1}} \left(...\right) \\ $$$$=\sum_{{p}=\mathrm{0}} ^{{n}} \:\:{C}_{\mathrm{2}{n}} ^{\mathrm{2}{p}} \:\:\:\mathrm{2}^{{p}+\mathrm{1}} \:+\mathrm{0}\:=\mathrm{2}\:\left\{\sum_{{p}=\mathrm{0}} ^{{n}} \:\:{C}_{\mathrm{2}{n}} ^{\mathrm{2}{p}} \:\:\mathrm{2}^{{p}} \:\right\}\:\:\Rightarrow{A}_{{n}} {is}\:{even}\:. \\ $$

Commented by Tawa1 last updated on 23/Apr/19

God bless you sir, i will check if i can understand the solution

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{will}\:\mathrm{check}\:\mathrm{if}\:\mathrm{i}\:\mathrm{can}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{solution} \\ $$

Commented by maxmathsup by imad last updated on 23/Apr/19

you are welcome sir, here i have used the binome formulae .

$${you}\:{are}\:{welcome}\:{sir},\:{here}\:{i}\:{have}\:{used}\:{the}\:{binome}\:{formulae}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com