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Question-10108

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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
fromVieta′sfomula:{x1+x2=2x1x2=−5x1n+2+x2n+2=(x1+x2)(x1n+1+x2n+1)−x1x2(x1n+x2n)=2(x1n+1+x2n+1)+5(x1n+x2n)whenn=2011:x12013+x22013=2(x12012+x22012)+5(x12011+x22011)=2B+5An=2012:x12014+x22014=2(x12013+x22013)+5(x12012+x22012)=2(2B+5A)+5B=9B+10An=2013:x12015+x22015=2(x12014+x22014)+5(x12013+x22013)=2(9B+10A)+5(2B+5A)=45A+28B
Commented by ridwan balatif last updated on 24/Jan/17
wow, now i know the solution, thank you so much sir
wow,nowiknowthesolution,thankyousomuchsir

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