find-the-next-4-term-and-the-n-th-term-1-2-5-26-

Question Number 28743 by Cheyboy last updated on 29/Jan/18

$${find}\:{the}\:{next}\:\mathrm{4}\:{term}\:{and}\:{the} \\ $$$${n}^{{th}} \:{term} \\ $$$$\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{26}……. \\ $$

Commented by Tinkutara last updated on 29/Jan/18

$${a}_{{n}} ={a}_{{n}−\mathrm{1}} ^{\mathrm{2}} +\mathrm{1} \\ $$

Commented by Cheyboy last updated on 29/Jan/18

$${thank}\:{u}\:{sir} \\ $$

Answered by ajfour last updated on 30/Jan/18

1 , 2 , 5, 26, 81, 186, 357, 610 1, 3, 21, 55, 105, 171, 253 2, 18, 34, 50, 66, 82 Let the first series 1,2,5,26,... be T series. middle one S series last one R series R_n =2+16(n−1) =16n−14 Σ_(n=1) ^n R_n =Σ_(n=1) ^n (S_(n+1) −S_n )=Σ_(n=1) ^n (16n−14) ⇒ S_(n+1) −S_1 =8n(n+1)−14n as S_1 =1 S_(n+1) =8n(n+1)−14n+1 S_(n+1) =T_(n+2) −T_(n+1) ⇒ Σ_(n=1) ^n (T_(n+2) −T_(n+1) )=Σ_(n=1) ^n [8n(n+1)−14n+1] =T_(n+2) −T_2 =((8n(n+1)(n+2))/3)−7n(n+1)+n as T_2 =2 T_(n+2) =((8n(n+1)(n+2))/3)−7n(n+1) +n+2 T_n =((8n(n−1)(n−2))/3) −7(n−1)(n−2)+n =(((n−1)(n−2)(8n−21))/3)+n .

$$\mathrm{1}\:\:\:,\:\:\:\mathrm{2}\:\:\:,\:\:\:\mathrm{5},\:\:\:\:\mathrm{26},\:\:\:\:\:\:\mathrm{81},\:\:\:\:\:\mathrm{186},\:\:\:\mathrm{357},\:\:\:\mathrm{610} \\ $$$$\:\:\:\:\mathrm{1},\:\:\:\:\:\:\:\mathrm{3},\:\:\:\:\:\:\mathrm{21},\:\:\:\:\:\mathrm{55},\:\:\:\:\:\mathrm{105},\:\:\mathrm{171},\:\:\:\mathrm{253} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2},\:\:\:\:\:\:\:\mathrm{18},\:\:\:\:\:\mathrm{34},\:\:\:\mathrm{50},\:\:\:\:\:\mathrm{66},\:\:\:\:\mathrm{82} \\ $$$${Let}\:{the}\:{first}\:{series}\:\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{26},… \\ $$$${be}\:{T}\:{series}. \\ $$$${middle}\:{one}\:{S}\:{series} \\ $$$${last}\:{one}\:\:{R}\:{series} \\ $$$${R}_{{n}} =\mathrm{2}+\mathrm{16}\left({n}−\mathrm{1}\right)\:=\mathrm{16}{n}−\mathrm{14} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{R}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({S}_{{n}+\mathrm{1}} −{S}_{{n}} \right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{16}{n}−\mathrm{14}\right) \\ $$$$\Rightarrow\:\:\:\:{S}_{{n}+\mathrm{1}} −{S}_{\mathrm{1}} =\mathrm{8}{n}\left({n}+\mathrm{1}\right)−\mathrm{14}{n} \\ $$$${as}\:{S}_{\mathrm{1}} =\mathrm{1} \\ $$$${S}_{{n}+\mathrm{1}} =\mathrm{8}{n}\left({n}+\mathrm{1}\right)−\mathrm{14}{n}+\mathrm{1} \\ $$$${S}_{{n}+\mathrm{1}} ={T}_{{n}+\mathrm{2}} −{T}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({T}_{{n}+\mathrm{2}} −{T}_{{n}+\mathrm{1}} \right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\mathrm{8}{n}\left({n}+\mathrm{1}\right)−\mathrm{14}{n}+\mathrm{1}\right] \\ $$$$={T}_{{n}+\mathrm{2}} −{T}_{\mathrm{2}} =\frac{\mathrm{8}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}−\mathrm{7}{n}\left({n}+\mathrm{1}\right)+{n} \\ $$$${as}\:\:{T}_{\mathrm{2}} =\mathrm{2} \\ $$$${T}_{{n}+\mathrm{2}} =\frac{\mathrm{8}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}−\mathrm{7}{n}\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{n}+\mathrm{2} \\ $$$$\boldsymbol{{T}}_{{n}} =\frac{\mathrm{8}\boldsymbol{{n}}\left(\boldsymbol{{n}}−\mathrm{1}\right)\left(\boldsymbol{{n}}−\mathrm{2}\right)}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{7}\left(\boldsymbol{{n}}−\mathrm{1}\right)\left(\boldsymbol{{n}}−\mathrm{2}\right)+\boldsymbol{{n}} \\ $$$$\:\:\:\:\:=\frac{\left(\boldsymbol{{n}}−\mathrm{1}\right)\left(\boldsymbol{{n}}−\mathrm{2}\right)\left(\mathrm{8}\boldsymbol{{n}}−\mathrm{21}\right)}{\mathrm{3}}+\boldsymbol{{n}}\:. \\ $$

Commented by Cheyboy last updated on 29/Jan/18