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Question Number 157795 by naka3546 last updated on 28/Oct/21
{a_n } is a natural number sequence for n ≥ 0 that satisfy recurrence relation a_(m+n) + a_(m−n) −m+n = 1 + (1/2) (a_(2m) +a_(2n) ) , for ∀ m,n nonnegative integers . Find a_(2016) .
$$\left\{{a}_{{n}} \right\}\:{is}\:{a}\:\:{natural}\:\:{number}\:\:{sequence}\:\:{for}\:\:{n}\:\geqslant\:\mathrm{0}\:\:{that}\:\:{satisfy} \\ $$$${recurrence}\:\:{relation}\:\:{a}_{{m}+{n}} \:+\:{a}_{{m}−{n}} \:−{m}+{n}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({a}_{\mathrm{2}{m}} \:+{a}_{\mathrm{2}{n}} \right)\:,\:\: \\ $$$${for}\:\:\forall\:{m},{n}\:\:{nonnegative}\:\:{integers}\:. \\ $$$${Find}\:\:{a}_{\mathrm{2016}} \:. \\ $$
Answered by Rasheed.Sindhi last updated on 30/Oct/21
a_(m+n) + a_(m−n) −m+n = 1 + (1/2) (a_(2m) +a_(2n) )_( A^ ) • determinant (((a_1 =3 (Given)))) • m=n:a_(2n) +a_0 =1+(1/2)(2a_(2n) )⇒a_0 =1 determinant (((a_0 =1))) ▶n=0:2a_m −m=1+(1/2)(a_(2m) +a_0 ) 1+(1/2)a_(2m) +(1/2)=2a_m −m (1/2)a_(2m) =2a_m −m−(3/2) a_(2m) =4a_m −2m−3 determinant ((( a_(2n) =4a_n −2n−3))) a_(2(2n)) =4a_(2n) −2n−3=4(4a_n −2n−3)−2n−3 a_(4n) =16a_n −10n−15 determinant (((a_(4n) =16a_n −10n−15))) Replacing n by 4n a_(4(4n)) =16a_(4n) −10n−15 a_(16n) =16(16a_n −10n−15)−10n−15 =256a_n −170n−255 Again replacing n by 2n: a_(16(2n)) =256a_(2n) −170(2n)−255 a_(32n) =256(4a_n −2n−3)−340n−255 =1024a_n −512n−768−340n−255 =1024a_n −852n−1023 determinant ((( determinant (((a_(32n) =1024a_n −852n−1023)))...A))) ▶A:(m=2n): a_(3n) +a_n −n=1+(1/2)(a_(4n) +a_(2n) ) 2a_(3n) +2a_n −2n=2+a_(4n) +a_(2n) 2a_(3n) +2a_n −2n−2=4a_(2n) −4n−3+4a_n −2n−3 =4(4a_n −2n−3)−4n−3+4a_n −2n−3 =16a_n −8n−12−4n−3+4a_n −2n−3 =20a_n −14n−18 2a_(3n) =18a_n −12n−16 a_(3n) =9a_n −6n−8 determinant (((a_(3n) =9a_n −6n−8))) Replacing n by 3n: a_(3(3n)) =9a_(3n) −6(3n)−8 a_(9n) =9(9a_n −6n−8)−18n−8 =81a_n −72n−80 determinant ((( determinant (((a_(9n) =81a_n −72n−80)))...B))) ▶A: { ((m→4n)),((n→3n)) :}⇒a_(7n) +a_n −7n+3n_( =1+(1/2)(a_(8n) +a_(6n) )) ▶ 2a_(7n) +2a_n −8n=2+a_(8n) +a_(6n) ▶2a_(7n) +2a_n −8n−2=a_(4(2n)) +a_(3(2n)) =16a_(2n) −10(2n)−15+9a_(2n) −6(2n)−8 =25a_(2n) −32n−23=25(4a_n −2n−3)−32n−23 =100a_n −50n−75−32n−23 =100a_n −82n−98 ▶2a_(7n) +2a_n −8n−2=100a_n −82n−98 2a_(7n) =98a_n −74n−96 determinant ((( determinant (((a_(7n) =49a_n −37n−48)))...C))) ▶Replacing n by 9n in A a_(32(9n)) =1024a_(9n) −852(9n)−1023 a_(288n) =1024(81a_n −72n−80)−7668n−1023 a_(288n) =82944a_n −81396n−82943 Again replacing n by 7n a_(288(7n)) =82944a_(7n) −81396(7n)−82943 a_(2016n) =82944(49a_n −37n−48)−569772n−82943 Now replacing n by 1 a_(2016n) =82944(49a_1 −37(1)−48)−569772(1)−82943 a_(2016n) =82944(49(3)−37−48)−569772−82943 =82944(62)−652715 determinant (( determinant (( determinant ((( a_(2016) =4489813)))))))
$$\underset{\:\:\:\:\:\overset{} {{A}}} {\underbrace{\:{a}_{{m}+{n}} \:+\:{a}_{{m}−{n}} \:−{m}+{n}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({a}_{\mathrm{2}{m}} \:+{a}_{\mathrm{2}{n}} \right)}} \\ $$$$\bullet\begin{array}{|c|}{{a}_{\mathrm{1}} =\mathrm{3}\:\left(\mathcal{G}{iven}\right)}\\\hline\end{array}\: \\ $$$$\bullet\:{m}={n}:{a}_{\mathrm{2}{n}} +{a}_{\mathrm{0}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{2}{n}} \right)\Rightarrow{a}_{\mathrm{0}} =\mathrm{1} \\ $$$$\:\:\:\begin{array}{|c|}{{a}_{\mathrm{0}} =\mathrm{1}}\\\hline\end{array} \\ $$$$\blacktriangleright{n}=\mathrm{0}:\mathrm{2}{a}_{{m}} −{m}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{2}{m}} +{a}_{\mathrm{0}} \right) \\ $$$$\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{\mathrm{2}{m}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}{a}_{{m}} −{m} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{a}_{\mathrm{2}{m}} =\mathrm{2}{a}_{{m}} −{m}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{a}_{\mathrm{2}{m}} =\mathrm{4}{a}_{{m}} −\mathrm{2}{m}−\mathrm{3} \\ $$$$\:\:\:\:\:\begin{array}{|c|}{\:{a}_{\mathrm{2}{n}} =\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}}\\\hline\end{array} \\ $$$${a}_{\mathrm{2}\left(\mathrm{2}{n}\right)} =\mathrm{4}{a}_{\mathrm{2}{n}} −\mathrm{2}{n}−\mathrm{3}=\mathrm{4}\left(\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}\right)−\mathrm{2}{n}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{4}{n}} =\mathrm{16}{a}_{{n}} −\mathrm{10}{n}−\mathrm{15} \\ $$$$\:\:\:\:\:\begin{array}{|c|}{{a}_{\mathrm{4}{n}} =\mathrm{16}{a}_{{n}} −\mathrm{10}{n}−\mathrm{15}}\\\hline\end{array} \\ $$$${Replacing}\:{n}\:{by}\:\mathrm{4}{n} \\ $$$${a}_{\mathrm{4}\left(\mathrm{4}{n}\right)} =\mathrm{16}{a}_{\mathrm{4}{n}} −\mathrm{10}{n}−\mathrm{15} \\ $$$${a}_{\mathrm{16}{n}} =\mathrm{16}\left(\mathrm{16}{a}_{{n}} −\mathrm{10}{n}−\mathrm{15}\right)−\mathrm{10}{n}−\mathrm{15} \\ $$$$\:\:\:\:\:\:\:=\mathrm{256}{a}_{{n}} −\mathrm{170}{n}−\mathrm{255} \\ $$$${Again}\:{replacing}\:{n}\:{by}\:\mathrm{2}{n}: \\ $$$${a}_{\mathrm{16}\left(\mathrm{2}{n}\right)} =\mathrm{256}{a}_{\mathrm{2}{n}} −\mathrm{170}\left(\mathrm{2}{n}\right)−\mathrm{255} \\ $$$${a}_{\mathrm{32}{n}} =\mathrm{256}\left(\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}\right)−\mathrm{340}{n}−\mathrm{255} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1024}{a}_{{n}} −\mathrm{512}{n}−\mathrm{768}−\mathrm{340}{n}−\mathrm{255} \\ $$$$\:\:=\mathrm{1024}{a}_{{n}} −\mathrm{852}{n}−\mathrm{1023} \\ $$$$\begin{array}{|c|}{\begin{array}{|c|}{{a}_{\mathrm{32}{n}} =\mathrm{1024}{a}_{{n}} −\mathrm{852}{n}−\mathrm{1023}}\\\hline\end{array}…{A}}\\\hline\end{array} \\ $$$$\blacktriangleright{A}:\left({m}=\mathrm{2}{n}\right):\:\:\:{a}_{\mathrm{3}{n}} +{a}_{{n}} −{n}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{4}{n}} +{a}_{\mathrm{2}{n}} \right) \\ $$$$\:\:\:\mathrm{2}{a}_{\mathrm{3}{n}} +\mathrm{2}{a}_{{n}} −\mathrm{2}{n}=\mathrm{2}+{a}_{\mathrm{4}{n}} +{a}_{\mathrm{2}{n}} \: \\ $$$$\mathrm{2}{a}_{\mathrm{3}{n}} +\mathrm{2}{a}_{{n}} −\mathrm{2}{n}−\mathrm{2}=\mathrm{4}{a}_{\mathrm{2}{n}} −\mathrm{4}{n}−\mathrm{3}+\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}\:\: \\ $$$$=\mathrm{4}\left(\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}\right)−\mathrm{4}{n}−\mathrm{3}+\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3} \\ $$$$=\mathrm{16}{a}_{{n}} −\mathrm{8}{n}−\mathrm{12}−\mathrm{4}{n}−\mathrm{3}+\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3} \\ $$$$=\mathrm{20}{a}_{{n}} −\mathrm{14}{n}−\mathrm{18} \\ $$$$\mathrm{2}{a}_{\mathrm{3}{n}} =\mathrm{18}{a}_{{n}} −\mathrm{12}{n}−\mathrm{16} \\ $$$${a}_{\mathrm{3}{n}} =\mathrm{9}{a}_{{n}} −\mathrm{6}{n}−\mathrm{8} \\ $$$$\:\begin{array}{|c|}{{a}_{\mathrm{3}{n}} =\mathrm{9}{a}_{{n}} −\mathrm{6}{n}−\mathrm{8}}\\\hline\end{array} \\ $$$${Replacing}\:{n}\:{by}\:\mathrm{3}{n}: \\ $$$$\:\:\:{a}_{\mathrm{3}\left(\mathrm{3}{n}\right)} =\mathrm{9}{a}_{\mathrm{3}{n}} −\mathrm{6}\left(\mathrm{3}{n}\right)−\mathrm{8} \\ $$$${a}_{\mathrm{9}{n}} =\mathrm{9}\left(\mathrm{9}{a}_{{n}} −\mathrm{6}{n}−\mathrm{8}\right)−\mathrm{18}{n}−\mathrm{8} \\ $$$$\:\:\:\:=\mathrm{81}{a}_{{n}} −\mathrm{72}{n}−\mathrm{80} \\ $$$$\begin{array}{|c|}{\begin{array}{|c|}{{a}_{\mathrm{9}{n}} =\mathrm{81}{a}_{{n}} −\mathrm{72}{n}−\mathrm{80}}\\\hline\end{array}…{B}}\\\hline\end{array} \\ $$$$\blacktriangleright{A}:\begin{cases}{{m}\rightarrow\mathrm{4}{n}}\\{{n}\rightarrow\mathrm{3}{n}}\end{cases}\Rightarrow\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{8}{n}} +{a}_{\mathrm{6}{n}} \right)} {{a}_{\mathrm{7}{n}} +{a}_{{n}} −\mathrm{7}{n}+\mathrm{3}{n}} \\ $$$$\blacktriangleright\:\:\mathrm{2}{a}_{\mathrm{7}{n}} +\mathrm{2}{a}_{{n}} −\mathrm{8}{n}=\mathrm{2}+{a}_{\mathrm{8}{n}} +{a}_{\mathrm{6}{n}} \\ $$$$\blacktriangleright\mathrm{2}{a}_{\mathrm{7}{n}} +\mathrm{2}{a}_{{n}} −\mathrm{8}{n}−\mathrm{2}={a}_{\mathrm{4}\left(\mathrm{2}{n}\right)} +{a}_{\mathrm{3}\left(\mathrm{2}{n}\right)} \\ $$$$=\mathrm{16}{a}_{\mathrm{2}{n}} −\mathrm{10}\left(\mathrm{2}{n}\right)−\mathrm{15}+\mathrm{9}{a}_{\mathrm{2}{n}} −\mathrm{6}\left(\mathrm{2}{n}\right)−\mathrm{8} \\ $$$$=\mathrm{25}{a}_{\mathrm{2}{n}} −\mathrm{32}{n}−\mathrm{23}=\mathrm{25}\left(\mathrm{4}{a}_{{n}} −\mathrm{2}{n}−\mathrm{3}\right)−\mathrm{32}{n}−\mathrm{23} \\ $$$$=\mathrm{100}{a}_{{n}} −\mathrm{50}{n}−\mathrm{75}−\mathrm{32}{n}−\mathrm{23} \\ $$$$=\mathrm{100}{a}_{{n}} −\mathrm{82}{n}−\mathrm{98} \\ $$$$\blacktriangleright\mathrm{2}{a}_{\mathrm{7}{n}} +\mathrm{2}{a}_{{n}} −\mathrm{8}{n}−\mathrm{2}=\mathrm{100}{a}_{{n}} −\mathrm{82}{n}−\mathrm{98} \\ $$$$\:\:\:\mathrm{2}{a}_{\mathrm{7}{n}} =\mathrm{98}{a}_{{n}} −\mathrm{74}{n}−\mathrm{96} \\ $$$$\begin{array}{|c|}{\begin{array}{|c|}{{a}_{\mathrm{7}{n}} =\mathrm{49}{a}_{{n}} −\mathrm{37}{n}−\mathrm{48}}\\\hline\end{array}…{C}}\\\hline\end{array} \\ $$$$\blacktriangleright{Replacing}\:{n}\:{by}\:\mathrm{9}{n}\:{in}\:{A} \\ $$$$\:\:\:{a}_{\mathrm{32}\left(\mathrm{9}{n}\right)} =\mathrm{1024}{a}_{\mathrm{9}{n}} −\mathrm{852}\left(\mathrm{9}{n}\right)−\mathrm{1023} \\ $$$$\:\:{a}_{\mathrm{288}{n}} =\mathrm{1024}\left(\mathrm{81}{a}_{{n}} −\mathrm{72}{n}−\mathrm{80}\right)−\mathrm{7668}{n}−\mathrm{1023} \\ $$$$\:{a}_{\mathrm{288}{n}} \:=\mathrm{82944}{a}_{{n}} −\mathrm{81396}{n}−\mathrm{82943} \\ $$$${Again}\:{replacing}\:{n}\:{by}\:\mathrm{7}{n}\: \\ $$$$\:{a}_{\mathrm{288}\left(\mathrm{7}{n}\right)} \:=\mathrm{82944}{a}_{\mathrm{7}{n}} −\mathrm{81396}\left(\mathrm{7}{n}\right)−\mathrm{82943} \\ $$$${a}_{\mathrm{2016}{n}} =\mathrm{82944}\left(\mathrm{49}{a}_{{n}} −\mathrm{37}{n}−\mathrm{48}\right)−\mathrm{569772}{n}−\mathrm{82943} \\ $$$$\:{Now}\:{replacing}\:{n}\:{by}\:\mathrm{1} \\ $$$${a}_{\mathrm{2016}{n}} =\mathrm{82944}\left(\mathrm{49}{a}_{\mathrm{1}} −\mathrm{37}\left(\mathrm{1}\right)−\mathrm{48}\right)−\mathrm{569772}\left(\mathrm{1}\right)−\mathrm{82943} \\ $$$${a}_{\mathrm{2016}{n}} =\mathrm{82944}\left(\mathrm{49}\left(\mathrm{3}\right)−\mathrm{37}−\mathrm{48}\right)−\mathrm{569772}−\mathrm{82943} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{82944}\left(\mathrm{62}\right)−\mathrm{652715} \\ $$$$\:\:\:\begin{array}{|c|}{\begin{array}{|c|}{\begin{array}{|c|}{\:{a}_{\mathrm{2016}} =\mathrm{4489813}}\\\hline\end{array}}\\\hline\end{array}}\\\hline\end{array} \\ $$
Commented by naka3546 last updated on 29/Oct/21
I′m so sorry, sir. I forgot to write that a_1 = 3 .
$${I}'{m}\:\:{so}\:\:{sorry},\:{sir}.\:\:{I}\:\:{forgot}\:\:{to}\:\:{write}\:\:{that}\:\:\:{a}_{\mathrm{1}} \:=\:\mathrm{3}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 30/Oct/21
naka sir, the answer is complete now.please verify it.If it′s not correct then please check my answer for mistakes. The approach is certainly right.
$${naka}\:{sir},\:{the}\:{answer}\:{is}\:{complete} \\ $$$${now}.{please}\:{verify}\:{it}.{If}\:{it}'{s}\:{not} \\ $$$${correct}\:{then}\:{please}\:{check}\:{my} \\ $$$${answer}\:{for}\:{mistakes}.\:\mathcal{T}{he}\:{approach} \\ $$$${is}\:{certainly}\:{right}. \\ $$
Commented by naka3546 last updated on 30/Oct/21
yes, it′s correct, sir.
$${yes},\:{it}'{s}\:\:{correct},\:\:{sir}.\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 30/Oct/21
Thank You!
$$\mathbb{T}\mathrm{han}\Bbbk\:\mathbb{Y}\mathrm{ou}! \\ $$

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