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Question Number 3061 by Rasheed Soomro last updated on 04/Dec/15
Determine nth term of the following sequence 1,0,−1,0,7,28,79,...
$$\mathrm{Determine}\:\:\mathrm{nth}\:\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sequence} \\ $$$$\mathrm{1},\mathrm{0},−\mathrm{1},\mathrm{0},\mathrm{7},\mathrm{28},\mathrm{79},… \\ $$
Commented by prakash jain last updated on 04/Dec/15
one techinque is to define define a series by successive subtraction until you get a known pattern. a_1 =1 a_2 =0 a_3 =−1 a_4 =0 a_5 =7 a_6 =28 a_7 =79 b_2 =a_2 −a_1 =−1 b_3 =−1 b_4 =1, b_5 =7, b_6 =21,b_7 =51 similary c_n =b_n −b_(n−1) c_3 =0 c_4 =2 c_5 =6 c_6 =14 c_7 =30 d_n =c_n −c_(n−1) d_4 =2 d_5 =4 d_6 =8 d_7 =16 d_n =2^(n−3) (n≥4) c_n =c_(n−1) +d_n (n≥4) b_n =b_(n−1) +c_n (n≥3) a_n =a_(n−1) +b_n (n≥2) for 8^(th) tem d_8 =32 c_8 =c_7 +d_8 =30+32=62 b_8 =b_7 +c_8 =51+62=113 a_8 =a_7 +b_8 =79+113=192
$${one}\:{techinque}\:{is}\:{to}\:{define} \\ $$$${define}\:{a}\:{series}\:{by}\:{successive}\:{subtraction} \\ $$$${until}\:{you}\:{get}\:{a}\:{known}\:{pattern}. \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{0} \\ $$$${a}_{\mathrm{3}} =−\mathrm{1} \\ $$$${a}_{\mathrm{4}} =\mathrm{0} \\ $$$${a}_{\mathrm{5}} =\mathrm{7} \\ $$$${a}_{\mathrm{6}} =\mathrm{28} \\ $$$${a}_{\mathrm{7}} =\mathrm{79} \\ $$$${b}_{\mathrm{2}} ={a}_{\mathrm{2}} −{a}_{\mathrm{1}} =−\mathrm{1} \\ $$$${b}_{\mathrm{3}} =−\mathrm{1} \\ $$$${b}_{\mathrm{4}} =\mathrm{1},\:{b}_{\mathrm{5}} =\mathrm{7},\:{b}_{\mathrm{6}} =\mathrm{21},{b}_{\mathrm{7}} =\mathrm{51} \\ $$$${similary}\:{c}_{{n}} ={b}_{{n}} −{b}_{{n}−\mathrm{1}} \\ $$$${c}_{\mathrm{3}} =\mathrm{0}\:\:{c}_{\mathrm{4}} =\mathrm{2}\:{c}_{\mathrm{5}} =\mathrm{6}\:{c}_{\mathrm{6}} =\mathrm{14}\:{c}_{\mathrm{7}} =\mathrm{30} \\ $$$${d}_{{n}} ={c}_{{n}} −{c}_{{n}−\mathrm{1}} \\ $$$${d}_{\mathrm{4}} =\mathrm{2}\:{d}_{\mathrm{5}} =\mathrm{4}\:{d}_{\mathrm{6}} =\mathrm{8}\:{d}_{\mathrm{7}} =\mathrm{16} \\ $$$${d}_{{n}} =\mathrm{2}^{{n}−\mathrm{3}} \:\:\left({n}\geqslant\mathrm{4}\right) \\ $$$${c}_{{n}} ={c}_{{n}−\mathrm{1}} +{d}_{{n}} \left({n}\geqslant\mathrm{4}\right) \\ $$$${b}_{{n}} ={b}_{{n}−\mathrm{1}} +{c}_{{n}} \left({n}\geqslant\mathrm{3}\right) \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +{b}_{{n}} \left({n}\geqslant\mathrm{2}\right) \\ $$$${for}\:\mathrm{8}^{{th}} \:{tem} \\ $$$${d}_{\mathrm{8}} =\mathrm{32} \\ $$$${c}_{\mathrm{8}} ={c}_{\mathrm{7}} +{d}_{\mathrm{8}} =\mathrm{30}+\mathrm{32}=\mathrm{62} \\ $$$${b}_{\mathrm{8}} ={b}_{\mathrm{7}} +{c}_{\mathrm{8}} =\mathrm{51}+\mathrm{62}=\mathrm{113} \\ $$$${a}_{\mathrm{8}} ={a}_{\mathrm{7}} +{b}_{\mathrm{8}} =\mathrm{79}+\mathrm{113}=\mathrm{192} \\ $$
Commented by Rasheed Soomro last updated on 05/Dec/15
THankS ^(for) Guidance^(Valueable) !
$$\mathcal{TH}{ank}\mathcal{S}\:\:\:^{{for}} \:\:\:\overset{\mathcal{V}{alueable}} {\mathcal{G}{uidance}}\:! \\ $$
Answered by Filup last updated on 04/Dec/15
1=2−1 ⇒ 2^1 −1 0=4−4 ⇒ 2^2 −2^2 −1=8−9 ⇒ 2^3 −3^2 etc. T_n =2^n −n^2
$$\mathrm{1}=\mathrm{2}−\mathrm{1}\:\:\Rightarrow\:\:\mathrm{2}^{\mathrm{1}} −\mathrm{1} \\ $$$$\mathrm{0}=\mathrm{4}−\mathrm{4}\:\:\Rightarrow\:\:\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \\ $$$$−\mathrm{1}=\mathrm{8}−\mathrm{9}\:\:\Rightarrow\:\:\mathrm{2}^{\mathrm{3}} −\mathrm{3}^{\mathrm{2}} \\ $$$${etc}. \\ $$$$ \\ $$$${T}_{{n}} =\mathrm{2}^{{n}} −{n}^{\mathrm{2}} \\ $$
Commented by Rasheed Soomro last updated on 04/Dec/15
V^(ery ) G^(OO) D!
$$\mathcal{V}\:^{{ery}\:} \mathcal{G}^{\mathcal{OO}} \mathcal{D}! \\ $$
Commented by RasheedAhmad last updated on 04/Dec/15
How did you reach that solution! Is there a way? I think these types of questions like ′ riddles ′ which can be solved only by guesses! Although intelligent guesses!
$$ \\ $$$$\mathcal{H}{ow}\:{did}\:{you}\:{reach}\:{that}\:{solution}! \\ $$$${Is}\:{there}\:{a}\:{way}?\:{I}\:{think}\:{these}\:{types} \\ $$$${of}\:{questions}\:{like}\:'\:{riddles}\:'\:{which} \\ $$$${can}\:{be}\:{solved}\:{only}\:{by}\:{guesses}!\: \\ $$$${Although}\:{intelligent}\:{guesses}! \\ $$
Commented by prakash jain last updated on 04/Dec/15
See comments in question about approach to find general terms. Uniqueness of formula is not guaranteed when only finite number of terms are known. Also see quesion 3003, where two different formulas can generate all known terms correctly but differ on next term.
$$\mathrm{See}\:\mathrm{comments}\:\mathrm{in}\:\mathrm{question}\:\mathrm{about}\:\mathrm{approach} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{general}\:\mathrm{terms}.\:\mathrm{Uniqueness}\:\mathrm{of} \\ $$$$\mathrm{formula}\:\mathrm{is}\:\mathrm{not}\:\mathrm{guaranteed}\:\mathrm{when}\:\mathrm{only}\:\mathrm{finite} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:\mathrm{are}\:\mathrm{known}. \\ $$$$\mathrm{Also}\:\mathrm{see}\:\mathrm{quesion}\:\mathrm{3003},\:\mathrm{where}\:\mathrm{two}\:\mathrm{different} \\ $$$$\mathrm{formulas}\:\mathrm{can}\:\mathrm{generate}\:\mathrm{all}\:\mathrm{known}\:\mathrm{terms} \\ $$$$\mathrm{correctly}\:\mathrm{but}\:\mathrm{differ}\:\mathrm{on}\:\mathrm{next}\:\mathrm{term}. \\ $$

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