Question Number 101474 by 175 last updated on 02/Jul/20
Answered by 1549442205 last updated on 07/Jul/20
![(see the solution of the question 101247) we known that [(2+(√2))^n ]=(2+(√2))^n +(2−(√2))^n −1 is an odd number.Putting A_n =(2+(√2))^n +(2−(√2))^n then a_n =A_n −1.Using induction method +For n=2 ⇒a_2 =[(2+(√2))^2 ]=[6+4(√2)]=11 =2.3+1+(3+1)=2a_1 +1+(a_1 +1) ⇒Equality true +Suppose Equality is true ∀n≤k which means we had a_k =2a_(k−1) +1+𝚺_(r=1) ^(k−1) (a_r +1) +Then a_(k+1) =[(2+(√2))^(k+1) ]=A_(k+1) −1 =(2+(√2))^(k+1) +(2−(√2))^(k+1) −1=[(2+(√2))^k +(2−(√2))^k ][(2+(√2) )+(2−(√2))] −2(2+(√2))^(k−1) −2(2−(√2))^(k−1) −1 =4A_k −2A_(k−1) −1=4(A_k −1)−2(A_(k−1) −1)+1 =4a_k −2a_(k−1) +1=2a_k +1+a_k +(a_k −2a_(k−1) ) =2a_k +1+a_k +{[2a_(k−1) +1+𝚺_(r=1) ^(k−1) (a_r +1)]−2a_(k−1) } =2a_k +1+(a_k +1+𝚺_(r=1) ^(k−1) (a_r +1)) =2a_k +1+𝚺_(r=1) ^(k) (a_r +1).This shows that Equality is also true for n=k+1.Hence, by induction Equality is true ∀n∈N,n>1 (q.e.d)](https://www.tinkutara.com/question/Q101584.png)
$$\left(\boldsymbol{\mathrm{see}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{question}}\:\mathrm{101247}\right) \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{known}}\:\boldsymbol{\mathrm{that}}\:\left[\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} \right]=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} −\mathrm{1} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{odd}}\:\boldsymbol{\mathrm{number}}.\boldsymbol{\mathrm{Putting}}\:\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{n}}} =\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} =\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{n}}} −\mathrm{1}.\boldsymbol{\mathrm{Using}}\:\boldsymbol{\mathrm{induction}}\:\boldsymbol{\mathrm{method}} \\ $$$$+\boldsymbol{\mathrm{For}}\:\boldsymbol{\mathrm{n}}=\mathrm{2}\:\Rightarrow\boldsymbol{\mathrm{a}}_{\mathrm{2}} =\left[\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right]=\left[\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right]=\mathrm{11} \\ $$$$=\mathrm{2}.\mathrm{3}+\mathrm{1}+\left(\mathrm{3}+\mathrm{1}\right)=\mathrm{2}\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\mathrm{1}+\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\mathrm{1}\right) \\ $$$$\Rightarrow\boldsymbol{\mathrm{Equality}}\:\boldsymbol{\mathrm{true}} \\ $$$$+\boldsymbol{\mathrm{Suppose}}\:\boldsymbol{\mathrm{Equality}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}\:\forall\boldsymbol{\mathrm{n}}\leqslant\boldsymbol{\mathrm{k}}\:\boldsymbol{\mathrm{which}} \\ $$$$\boldsymbol{\mathrm{means}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{had}}\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} =\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} +\mathrm{1}+\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{k}−\mathrm{1}} {\boldsymbol{\Sigma}}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{r}}} +\mathrm{1}\right) \\ $$$$+\boldsymbol{\mathrm{Then}}\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}+\mathrm{1}} =\left[\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} \right]=\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{k}}+\mathrm{1}} −\mathrm{1} \\ $$$$=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} −\mathrm{1}=\left[\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{k}}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{k}}} \right]\left[\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right)+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\right] \\ $$$$−\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{k}−\mathrm{1}} −\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\mathrm{k}−\mathrm{1}} −\mathrm{1} \\ $$$$=\mathrm{4}\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{k}}} −\mathrm{2}\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} −\mathrm{1}=\mathrm{4}\left(\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{k}}} −\mathrm{1}\right)−\mathrm{2}\left(\boldsymbol{\mathrm{A}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} −\mathrm{1}\right)+\mathrm{1} \\ $$$$=\mathrm{4}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} −\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} +\mathrm{1}=\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\mathrm{1}+\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} −\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} \right) \\ $$$$=\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\mathrm{1}+\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\left\{\left[\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} +\mathrm{1}+\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{k}−\mathrm{1}} {\boldsymbol{\Sigma}}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{r}}} +\mathrm{1}\right)\right]−\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}−\mathrm{1}} \right\} \\ $$$$=\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\mathrm{1}+\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\mathrm{1}+\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{k}−\mathrm{1}} {\boldsymbol{\Sigma}}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{r}}} +\mathrm{1}\right)\right) \\ $$$$=\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} +\mathrm{1}+\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{k}} {\boldsymbol{\Sigma}}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{r}}} +\mathrm{1}\right).\boldsymbol{\mathrm{This}}\:\boldsymbol{\mathrm{shows}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{Equality}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{also}}\:\boldsymbol{\mathrm{true}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{n}}=\boldsymbol{\mathrm{k}}+\mathrm{1}.\boldsymbol{\mathrm{Hence}}, \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{induction}}\:\boldsymbol{\mathrm{Equality}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}\:\forall\boldsymbol{\mathrm{n}}\in\boldsymbol{\mathrm{N}},\boldsymbol{\mathrm{n}}>\mathrm{1} \\ $$$$\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$