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Question Number 129403 by bemath last updated on 15/Jan/21
$$\:\mathrm{f}\left(\mathrm{n}\right)\:=\:\mathrm{4f}\left(\mathrm{n}−\mathrm{1}\right)\:−\mathrm{4f}\left(\mathrm{n}−\mathrm{2}\right)\:+\:\mathrm{n}^{\mathrm{2}} \: \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{2}\:,\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}\: \\ $$
Commented by talminator2856791 last updated on 15/Jan/21
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}? \\ $$
Answered by mr W last updated on 15/Jan/21
![let f_n =g_n +An^2 +Bn+C g_n +An^2 +Bn+C=4g_(n−1) −4g_(n−2) +4A[(n−1)^2 −(n−2)^2 ]+4B[(n−1)−(n−2)]+4(C−C)+n^2 g_n +An^2 +Bn+C=4g_(n−1) −4g_(n−2) +8An−12A+4B+n^2 ⇒A=1 ⇒B=8 ⇒C=20 g_n −4g_(n−1) +4g_(n−2) =0 t^2 −4t+4=0 (t−2)^2 =0 ⇒t=2 g_n =(α+βn)2^n ⇒f(n)=(α+βn)2^n +n^2 +8n+20 f(0)=α+20=2 ⇒α=−18 f(1)=(−18+β)2+1+8+20=5 ⇒β=6 ⇒f(n)=6(n−3)2^n +n^2 +8n+20](Q129434.png)
$${let}\:{f}_{{n}} ={g}_{{n}} +{An}^{\mathrm{2}} +{Bn}+{C} \\ $$$${g}_{{n}} +{An}^{\mathrm{2}} +{Bn}+{C}=\mathrm{4}{g}_{{n}−\mathrm{1}} −\mathrm{4}{g}_{{n}−\mathrm{2}} +\mathrm{4}{A}\left[\left({n}−\mathrm{1}\right)^{\mathrm{2}} −\left({n}−\mathrm{2}\right)^{\mathrm{2}} \right]+\mathrm{4}{B}\left[\left({n}−\mathrm{1}\right)−\left({n}−\mathrm{2}\right)\right]+\mathrm{4}\left({C}−{C}\right)+{n}^{\mathrm{2}} \\ $$$${g}_{{n}} +{An}^{\mathrm{2}} +{Bn}+{C}=\mathrm{4}{g}_{{n}−\mathrm{1}} −\mathrm{4}{g}_{{n}−\mathrm{2}} +\mathrm{8}{An}−\mathrm{12}{A}+\mathrm{4}{B}+{n}^{\mathrm{2}} \\ $$$$\Rightarrow{A}=\mathrm{1} \\ $$$$\Rightarrow{B}=\mathrm{8} \\ $$$$\Rightarrow{C}=\mathrm{20} \\ $$$${g}_{{n}} −\mathrm{4}{g}_{{n}−\mathrm{1}} +\mathrm{4}{g}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{2} \\ $$$${g}_{{n}} =\left(\alpha+\beta{n}\right)\mathrm{2}^{{n}} \\ $$$$\Rightarrow{f}\left({n}\right)=\left(\alpha+\beta{n}\right)\mathrm{2}^{{n}} +{n}^{\mathrm{2}} +\mathrm{8}{n}+\mathrm{20} \\ $$$${f}\left(\mathrm{0}\right)=\alpha+\mathrm{20}=\mathrm{2}\:\Rightarrow\alpha=−\mathrm{18} \\ $$$${f}\left(\mathrm{1}\right)=\left(−\mathrm{18}+\beta\right)\mathrm{2}+\mathrm{1}+\mathrm{8}+\mathrm{20}=\mathrm{5}\:\Rightarrow\beta=\mathrm{6} \\ $$$$\Rightarrow{f}\left({n}\right)=\mathrm{6}\left({n}−\mathrm{3}\right)\mathrm{2}^{{n}} +{n}^{\mathrm{2}} +\mathrm{8}{n}+\mathrm{20} \\ $$