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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
	$from Vieta′s fomula: { ((x_1 +x_2 =2)),((x_1 x_2 =−5)) :} x_1 ^(n+2) +x_2 ^(n+2) =(x_1 +x_2 )(x_1 ^(n+1) +x_2 ^(n+1) )−x_1 x_2 (x_1 ^n +x_2 ^n ) =2(x_1 ^(n+1) +x_2 ^(n+1) )+5(x_1 ^n +x_2 ^n ) when n=2011: x_1 ^(2013) +x_2 ^(2013) =2(x_1 ^(2012) +x_2 ^(2012) )+5(x_1 ^(2011) +x_2 ^(2011) ) =2B+5A n=2012: x_1 ^(2014) +x_2 ^(2014) =2(x_1 ^(2013) +x_2 ^(2013) )+5(x_1 ^(2012) +x_2 ^(2012) ) =2(2B+5A)+5B =9B+10A n=2013: x_1 ^(2015) +x_2 ^(2015) =2(x_1 ^(2014) +x_2 ^(2014) )+5(x_1 ^(2013) +x_2 ^(2013) ) =2(9B+10A)+5(2B+5A) =45A+28B$
	$f r o m {V i e t a}^{'} s f o m u l a :$ ${\begin{cases} x_{1} + x_{2} = 2 \\ x_{1} x_{2} = - 5 \end{cases}$ $x_{1}^{n + 2} + x_{2}^{n + 2} = (x_{1} + x_{2}) (x_{1}^{n + 1} + x_{2}^{n + 1}) - x_{1} x_{2} (x_{1}^{n} + x_{2}^{n})$ $= 2 (x_{1}^{n + 1} + x_{2}^{n + 1}) + 5 (x_{1}^{n} + x_{2}^{n})$ $w h e n n = 2011 :$ $x_{1}^{2013} + x_{2}^{2013} = 2 (x_{1}^{2012} + x_{2}^{2012}) + 5 (x_{1}^{2011} + x_{2}^{2011})$ $= 2 B + 5 A$ $n = 2012 :$ $x_{1}^{2014} + x_{2}^{2014} = 2 (x_{1}^{2013} + x_{2}^{2013}) + 5 (x_{1}^{2012} + x_{2}^{2012})$ $= 2 (2 B + 5 A) + 5 B$ $= 9 B + 10 A$ $n = 2013 :$ $x_{1}^{2015} + x_{2}^{2015} = 2 (x_{1}^{2014} + x_{2}^{2014}) + 5 (x_{1}^{2013} + x_{2}^{2013})$ $= 2 (9 B + 10 A) + 5 (2 B + 5 A)$ $= 45 A + 28 B$
	Commented by ridwan balatif last updated on 24/Jan/17

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