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

Commented byridwan balatif last updated on 24/Jan/17

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