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Question Number 64126 by Cheyboy last updated on 13/Jul/19

Answered by ajfour last updated on 13/Jul/19

  32           0          −8         −10              −24           −32     −8         −2                −14                     24          6              −12                          −18       −18  d_n =−18  c_n =24−18(n−1)  b_(n+1) −b_n = c_n   Σ_(n=1) ^n (b_(n+1) −b_n )=Σ_(n=1) ^n c_n   ⇒   b_(n+1) −b_1 =24n−9n(n−1)  ⇒   b_(n+1) =33n−9n^2 −32  Σ_(n=1) ^n (a_(n+2) −a_(n+1) )=Σ_(n=1) ^n b_(n+1)   a_(n+2) −a_2 =((33n(n+1))/2)−((9n(n+1)(2n+1))/6)−32n  a_n =((33(n−2)(n−1))/2)−((3(n−2)(n−1)(2n−3))/2)−32(n−2)     =3(n−1)(n−2)(7−n)−32(n−2)     = 3(−n^3 +10n^2 −23n+14)−32n+64  ________________________  a_n =−3n^3 +30n^2 −101n+106  ________________________

$$\:\:\mathrm{32}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:−\mathrm{32}\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24}\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{18}\:\:\:\:\:\:\:−\mathrm{18} \\ $$$${d}_{{n}} =−\mathrm{18} \\ $$$${c}_{{n}} =\mathrm{24}−\mathrm{18}\left({n}−\mathrm{1}\right) \\ $$$${b}_{{n}+\mathrm{1}} −{b}_{{n}} =\:{c}_{{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({b}_{{n}+\mathrm{1}} −{b}_{{n}} \right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{c}_{{n}} \\ $$$$\Rightarrow\:\:\:{b}_{{n}+\mathrm{1}} −{b}_{\mathrm{1}} =\mathrm{24}{n}−\mathrm{9}{n}\left({n}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:{b}_{{n}+\mathrm{1}} =\mathrm{33}{n}−\mathrm{9}{n}^{\mathrm{2}} −\mathrm{32} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} \right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{b}_{{n}+\mathrm{1}} \\ $$$${a}_{{n}+\mathrm{2}} −{a}_{\mathrm{2}} =\frac{\mathrm{33}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{9}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\mathrm{32}{n} \\ $$$${a}_{{n}} =\frac{\mathrm{33}\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{3}\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)}{\mathrm{2}}−\mathrm{32}\left({n}−\mathrm{2}\right) \\ $$$$\:\:\:=\mathrm{3}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left(\mathrm{7}−{n}\right)−\mathrm{32}\left({n}−\mathrm{2}\right) \\ $$$$\:\:\:=\:\mathrm{3}\left(−{n}^{\mathrm{3}} +\mathrm{10}{n}^{\mathrm{2}} −\mathrm{23}{n}+\mathrm{14}\right)−\mathrm{32}{n}+\mathrm{64} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${a}_{{n}} =−\mathrm{3}{n}^{\mathrm{3}} +\mathrm{30}{n}^{\mathrm{2}} −\mathrm{101}{n}+\mathrm{106} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

Commented by Cheyboy last updated on 13/Jul/19

Thank you sir

$${Thank}\:{you}\:{sir} \\ $$

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