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a-1-a-2-a-3-a-n-is-a-sequence-satifies-that-a-n-2-a-n-1-a-n-for-n-1-suppose-the-sum-of-the-first-999-terms-1003-and-the-sum-of-the-first-1003-terms-999-find-the-




Question Number 192597 by York12 last updated on 22/May/23
a_(1 )  , a_2  , a_(3 )  , .... , a_n  is a sequence satifies that  a_(n+2) =a_(n+1) −a_n  for n ≥ 1. suppose the sum   of the first 999 terms = 1003 and the sum  of the first 1003 terms = −999 find the   sum of the first 2002 terms.
$${a}_{\mathrm{1}\:} \:,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}\:} \:,\:….\:,\:{a}_{{n}} \:{is}\:{a}\:{sequence}\:{satifies}\:{that} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:{for}\:{n}\:\geq\:\mathrm{1}.\:{suppose}\:{the}\:{sum}\: \\ $$$${of}\:{the}\:{first}\:\mathrm{999}\:{terms}\:=\:\mathrm{1003}\:{and}\:{the}\:{sum} \\ $$$${of}\:{the}\:{first}\:\mathrm{1003}\:{terms}\:=\:−\mathrm{999}\:{find}\:{the}\: \\ $$$${sum}\:{of}\:{the}\:{first}\:\mathrm{2002}\:{terms}. \\ $$
Answered by MM42 last updated on 22/May/23
a_4 =a_3 −a_2 =−a_1   &  a_5 =a_4 −a_3 =−a_2    &  a_6 =a_5 −a_4 =−a_3   ⇒a_(6k+1) =a_1  & a_(6k+2) =a_2  & a_(6k+3) =a_3   a_(6k+4) =−a_1  & a_(6k+5) =−a_2  & a_(6k) =−a_3   999=6×166+3⇒S_(999) =a_1 +a_2 +a_3 =2a_2 =1003⇒a_2 =501.5  1003=6×167+1⇒S_(1003) =a_1 ⇒a_1 =−999  ⇒a_3 =1500.5  2002=6×336+4⇒S_(2002) =a_1 +a_2 +a_3 +a_4 =a_2 +a_3 =2002
$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} =−{a}_{\mathrm{1}} \:\:\&\:\:{a}_{\mathrm{5}} ={a}_{\mathrm{4}} −{a}_{\mathrm{3}} =−{a}_{\mathrm{2}} \:\:\:\&\:\:{a}_{\mathrm{6}} ={a}_{\mathrm{5}} −{a}_{\mathrm{4}} =−{a}_{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{6}{k}+\mathrm{1}} ={a}_{\mathrm{1}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{2}} ={a}_{\mathrm{2}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{3}} ={a}_{\mathrm{3}} \\ $$$${a}_{\mathrm{6}{k}+\mathrm{4}} =−{a}_{\mathrm{1}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{5}} =−{a}_{\mathrm{2}} \:\&\:{a}_{\mathrm{6}{k}} =−{a}_{\mathrm{3}} \\ $$$$\mathrm{999}=\mathrm{6}×\mathrm{166}+\mathrm{3}\Rightarrow{S}_{\mathrm{999}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} =\mathrm{2}{a}_{\mathrm{2}} =\mathrm{1003}\Rightarrow{a}_{\mathrm{2}} =\mathrm{501}.\mathrm{5} \\ $$$$\mathrm{1003}=\mathrm{6}×\mathrm{167}+\mathrm{1}\Rightarrow{S}_{\mathrm{1003}} ={a}_{\mathrm{1}} \Rightarrow{a}_{\mathrm{1}} =−\mathrm{999} \\ $$$$\Rightarrow{a}_{\mathrm{3}} =\mathrm{1500}.\mathrm{5} \\ $$$$\mathrm{2002}=\mathrm{6}×\mathrm{336}+\mathrm{4}\Rightarrow{S}_{\mathrm{2002}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +{a}_{\mathrm{4}} ={a}_{\mathrm{2}} +{a}_{\mathrm{3}} =\mathrm{2002} \\ $$
Commented by York12 last updated on 22/May/23
a_2 =501.5  s_(2002) =999+2×501.5=2002
$${a}_{\mathrm{2}} =\mathrm{501}.\mathrm{5} \\ $$$${s}_{\mathrm{2002}} =\mathrm{999}+\mathrm{2}×\mathrm{501}.\mathrm{5}=\mathrm{2002} \\ $$$$ \\ $$
Commented by York12 last updated on 22/May/23
    a_4 =a_3 −a_2 =−a_1   &  a_5 =a_4 −a_3 =−a_2    &  a_6 =a_5 −a_4 =−a_3   ⇒a_(6k+1) = a_1  & a_(6k+2) = a_2  & a_(6k+3) = a_3   a_(6k+4) = −a_1  & a_(6k+5) =−a_2  & a_(6k) = −a_3   999=6×166+3⇒S_(999) =a_1 +a_2 +a_3 =2a_2 =1003⇒a_2  = 501.5  1003=6×167+1⇒S_(1003) =a_1 ⇒a_1 =−999  ⇛a_3 = 1500.5  2002=6×336+4⇒S_(2002) =a_1 +a_2 +a_3 +a_4 =a_2 +a_3 =2002
$$ \\ $$$$ \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} =−{a}_{\mathrm{1}} \:\:\&\:\:{a}_{\mathrm{5}} ={a}_{\mathrm{4}} −{a}_{\mathrm{3}} =−{a}_{\mathrm{2}} \:\:\:\&\:\:{a}_{\mathrm{6}} ={a}_{\mathrm{5}} −{a}_{\mathrm{4}} =−{a}_{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{6}{k}+\mathrm{1}} =\:{a}_{\mathrm{1}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{2}} =\:{a}_{\mathrm{2}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{3}} =\:{a}_{\mathrm{3}} \\ $$$${a}_{\mathrm{6}{k}+\mathrm{4}} =\:−{a}_{\mathrm{1}} \:\&\:{a}_{\mathrm{6}{k}+\mathrm{5}} =−{a}_{\mathrm{2}} \:\&\:{a}_{\mathrm{6}{k}} =\:−{a}_{\mathrm{3}} \\ $$$$\mathrm{999}=\mathrm{6}×\mathrm{166}+\mathrm{3}\Rightarrow{S}_{\mathrm{999}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} =\mathrm{2}{a}_{\mathrm{2}} =\mathrm{1003}\Rightarrow{a}_{\mathrm{2}} \:=\:\mathrm{501}.\mathrm{5} \\ $$$$\mathrm{1003}=\mathrm{6}×\mathrm{167}+\mathrm{1}\Rightarrow{S}_{\mathrm{1003}} ={a}_{\mathrm{1}} \Rightarrow{a}_{\mathrm{1}} =−\mathrm{999} \\ $$$$\Rrightarrow{a}_{\mathrm{3}} =\:\mathrm{1500}.\mathrm{5} \\ $$$$\mathrm{2002}=\mathrm{6}×\mathrm{336}+\mathrm{4}\Rightarrow{S}_{\mathrm{2002}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +{a}_{\mathrm{4}} ={a}_{\mathrm{2}} +{a}_{\mathrm{3}} =\mathrm{2002} \\ $$$$ \\ $$

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