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Question Number 10108 by ridwan balatif last updated on 24/Jan/17

Answered by nume1114 last updated on 24/Jan/17

from Vieta′s fomula: { ((x_1 +x_2 =2)),((x_1 x_2 =−5)) :} x_1 ^(n+2) +x_2 ^(n+2) =(x_1 +x_2 )(x_1 ^(n+1) +x_2 ^(n+1) )−x_1 x_2 (x_1 ^n +x_2 ^n ) =2(x_1 ^(n+1) +x_2 ^(n+1) )+5(x_1 ^n +x_2 ^n ) when n=2011: x_1 ^(2013) +x_2 ^(2013) =2(x_1 ^(2012) +x_2 ^(2012) )+5(x_1 ^(2011) +x_2 ^(2011) ) =2B+5A n=2012: x_1 ^(2014) +x_2 ^(2014) =2(x_1 ^(2013) +x_2 ^(2013) )+5(x_1 ^(2012) +x_2 ^(2012) ) =2(2B+5A)+5B =9B+10A n=2013: x_1 ^(2015) +x_2 ^(2015) =2(x_1 ^(2014) +x_2 ^(2014) )+5(x_1 ^(2013) +x_2 ^(2013) ) =2(9B+10A)+5(2B+5A) =45A+28B

$${from}\:{Vieta}'{s}\:{fomula}: \\ $$$$\begin{cases}{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{2}}\\{{x}_{\mathrm{1}} {x}_{\mathrm{2}} =−\mathrm{5}}\end{cases} \\ $$$$ \\ $$$${x}_{\mathrm{1}} ^{{n}+\mathrm{2}} +{x}_{\mathrm{2}} ^{{n}+\mathrm{2}} =\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)\left({x}_{\mathrm{1}} ^{{n}+\mathrm{1}} +{x}_{\mathrm{2}} ^{{n}+\mathrm{1}} \right)−{x}_{\mathrm{1}} {x}_{\mathrm{2}} \left({x}_{\mathrm{1}} ^{{n}} +{x}_{\mathrm{2}} ^{{n}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left({x}_{\mathrm{1}} ^{{n}+\mathrm{1}} +{x}_{\mathrm{2}} ^{{n}+\mathrm{1}} \right)+\mathrm{5}\left({x}_{\mathrm{1}} ^{{n}} +{x}_{\mathrm{2}} ^{{n}} \right) \\ $$$${when}\:{n}=\mathrm{2011}: \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2013}} +{x}_{\mathrm{2}} ^{\mathrm{2013}} =\mathrm{2}\left({x}_{\mathrm{1}} ^{\mathrm{2012}} +{x}_{\mathrm{2}} ^{\mathrm{2012}} \right)+\mathrm{5}\left({x}_{\mathrm{1}} ^{\mathrm{2011}} +{x}_{\mathrm{2}} ^{\mathrm{2011}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{B}+\mathrm{5}{A} \\ $$$${n}=\mathrm{2012}: \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2014}} +{x}_{\mathrm{2}} ^{\mathrm{2014}} =\mathrm{2}\left({x}_{\mathrm{1}} ^{\mathrm{2013}} +{x}_{\mathrm{2}} ^{\mathrm{2013}} \right)+\mathrm{5}\left({x}_{\mathrm{1}} ^{\mathrm{2012}} +{x}_{\mathrm{2}} ^{\mathrm{2012}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{2}{B}+\mathrm{5}{A}\right)+\mathrm{5}{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}{B}+\mathrm{10}{A} \\ $$$${n}=\mathrm{2013}: \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2015}} +{x}_{\mathrm{2}} ^{\mathrm{2015}} =\mathrm{2}\left({x}_{\mathrm{1}} ^{\mathrm{2014}} +{x}_{\mathrm{2}} ^{\mathrm{2014}} \right)+\mathrm{5}\left({x}_{\mathrm{1}} ^{\mathrm{2013}} +{x}_{\mathrm{2}} ^{\mathrm{2013}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{9}{B}+\mathrm{10}{A}\right)+\mathrm{5}\left(\mathrm{2}{B}+\mathrm{5}{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{45}{A}+\mathrm{28}{B} \\ $$

Commented by ridwan balatif last updated on 24/Jan/17

wow, now i know the solution, thank you so much sir

$$\mathrm{wow},\:\mathrm{now}\:\mathrm{i}\:\mathrm{know}\:\mathrm{the}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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