Question Number 28012 by JI Siam last updated on 18/Jan/18
$$\mathrm{1},\mathrm{4},\mathrm{5},\mathrm{16},\mathrm{17},\mathrm{20}…….\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{68th}\:\mathrm{term}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sequnce}? \\ $$
Commented by JI Siam last updated on 18/Jan/18
$$\mathrm{please}\:\mathrm{give}\:\mathrm{answer}\:\mathrm{fast} \\ $$
Answered by ajfour last updated on 18/Jan/18
$${T}_{{r}+\mathrm{4}} =\mathrm{2}\left(−\mathrm{1}\right)^{{r}} −\frac{\mathrm{2}}{\mathrm{3}}{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}\left({r}+\mathrm{1}\right)+\mathrm{7}{r}+\mathrm{14} \\ $$$${T}_{\mathrm{68}} ={T}_{\mathrm{64}+\mathrm{4}} \\ $$$$\:\:\:\:\:=\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{64}×\mathrm{65}×\mathrm{66}\right)+\mathrm{64}×\mathrm{65} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{7}×\mathrm{64}+\mathrm{14} \\ $$$$\:\:\:\:\:=\mathrm{2}−\mathrm{43}×\mathrm{64}×\mathrm{65}+\mathrm{448}+\mathrm{14} \\ $$$$\:\:\:\:\:=−\mathrm{43}×\mathrm{64}×\mathrm{65}+\mathrm{464} \\ $$$$\:\:\:\:\:=−\mathrm{178416}\:. \\ $$
Commented by ajfour last updated on 19/Jan/18
![1 ,4, 5, 16, 17, 20, 5,...T series 3, 1, 11, 1, 3, −15,...S series −2, 10, −10, 2, −18,... R series 12, −20, 12,−20,... Q series −32, 32,−32,.. P series P_r =32(−1)^r Q_(r+1) −Q_r =P_r so Σ_(r=1) ^r (Q_(r+1) −Q_r )=Σ_(r=1) ^r 32(−1)^r ⇒ Q_(r+1) −Q_1 =16[(−1)^r −1] so Q_(r+1) −12=16(−1)^r −16 Q_(r+1) =16(−1)^r −4 proceeding in a similar manner Σ_(r=1) ^r (R_(r+2) −R_(r+1) )=Σ_(r=1) ^r Q_(r+1) R_(r+2) =8(−1)^r −4r+2 Then Σ_(r=1) ^r (S_(r+3) −S_(r+2) )=Σ_(r=1) ^r R_(r+2) S_(r+3) =4(−1)^r −2r(r+1)+2r+7 and finally, after another such step T_(r+4) =2(−1)^r −(2/3)r(r+1)(r+2)+ r(r+1)+7r+14 .](https://www.tinkutara.com/question/Q28029.png)
$$\mathrm{1}\:,\mathrm{4},\:\:\:\:\:\mathrm{5},\:\mathrm{16},\:\:\:\:\:\:\mathrm{17},\:\mathrm{20},\:\:\:\:\:\:\:\:\mathrm{5},…{T}\:{series} \\ $$$$\:\:\mathrm{3},\:\:\:\:\:\mathrm{1},\:\:\mathrm{11},\:\:\:\:\:\:\:\mathrm{1},\:\:\mathrm{3},\:\:\:−\mathrm{15},…{S}\:{series} \\ $$$$\:\:\:−\mathrm{2},\:\:\mathrm{10},\:−\mathrm{10},\:\:\mathrm{2},\:\:−\mathrm{18},…\:{R}\:{series} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{12},\:−\mathrm{20},\:\mathrm{12},−\mathrm{20},…\:{Q}\:{series} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{32},\:\:\:\:\mathrm{32},−\mathrm{32},..\:\:{P}\:{series} \\ $$$${P}_{{r}} =\mathrm{32}\left(−\mathrm{1}\right)^{{r}} \\ $$$${Q}_{{r}+\mathrm{1}} −{Q}_{{r}} ={P}_{{r}} \\ $$$${so}\:\:\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}\left({Q}_{{r}+\mathrm{1}} −{Q}_{{r}} \right)=\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}\mathrm{32}\left(−\mathrm{1}\right)^{{r}} \\ $$$$\Rightarrow\:{Q}_{{r}+\mathrm{1}} −{Q}_{\mathrm{1}} =\mathrm{16}\left[\left(−\mathrm{1}\right)^{{r}} −\mathrm{1}\right] \\ $$$${so}\:\:{Q}_{{r}+\mathrm{1}} −\mathrm{12}=\mathrm{16}\left(−\mathrm{1}\right)^{{r}} −\mathrm{16} \\ $$$${Q}_{{r}+\mathrm{1}} =\mathrm{16}\left(−\mathrm{1}\right)^{{r}} −\mathrm{4} \\ $$$${proceeding}\:{in}\:{a}\:{similar}\:{manner} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}\left({R}_{{r}+\mathrm{2}} −{R}_{{r}+\mathrm{1}} \right)=\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}{Q}_{{r}+\mathrm{1}} \\ $$$${R}_{{r}+\mathrm{2}} =\mathrm{8}\left(−\mathrm{1}\right)^{{r}} −\mathrm{4}{r}+\mathrm{2} \\ $$$${Then}\:\:\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}\left({S}_{{r}+\mathrm{3}} −{S}_{{r}+\mathrm{2}} \right)=\underset{{r}=\mathrm{1}} {\overset{{r}} {\sum}}{R}_{{r}+\mathrm{2}} \\ $$$${S}_{{r}+\mathrm{3}} =\mathrm{4}\left(−\mathrm{1}\right)^{{r}} −\mathrm{2}{r}\left({r}+\mathrm{1}\right)+\mathrm{2}{r}+\mathrm{7} \\ $$$${and}\:{finally},\:{after}\:{another}\:{such}\:{step} \\ $$$${T}_{{r}+\mathrm{4}} =\mathrm{2}\left(−\mathrm{1}\right)^{{r}} −\frac{\mathrm{2}}{\mathrm{3}}{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{r}\left({r}+\mathrm{1}\right)+\mathrm{7}{r}+\mathrm{14}\:. \\ $$