Question Number 159939 by Ar Brandon last updated on 22/Nov/21
$$\mathrm{U}_{{n}+\mathrm{2}} −\mathrm{2U}_{{n}+\mathrm{1}} +\mathrm{U}_{{n}} =\mathrm{800} \\ $$
Commented by mr W last updated on 22/Nov/21
$${two}\:{terms}\:{must}\:{be}\:{given},\:{for}\:{example} \\ $$$${U}_{\mathrm{0}} ,\:{U}_{\mathrm{1}} \:{or}\:{U}_{\mathrm{1}} ,\:{U}_{\mathrm{2}} . \\ $$
Commented by Ar Brandon last updated on 22/Nov/21
$$\mathrm{Ok}\:\mathrm{Sir}.\:\mathrm{Let}\:\mathrm{me}\:\mathrm{review}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{attention}. \\ $$
Answered by mr W last updated on 22/Nov/21
$${assume}\:{U}_{\mathrm{0}} =\mathrm{1},\:{U}_{\mathrm{1}} =\mathrm{2}. \\ $$$$\left({U}_{{n}+\mathrm{2}} −{U}_{{n}+\mathrm{1}} \right)−\left({U}_{{n}+\mathrm{1}} −{U}_{{n}} \right)=\mathrm{800} \\ $$$${let}\:{V}_{{n}} ={U}_{{n}+\mathrm{1}} −{U}_{{n}} \\ $$$${V}_{{n}+\mathrm{1}} −{V}_{{n}} =\mathrm{800} \\ $$$${this}\:{is}\:{an}\:{A}.{P}.\:{with}\:{common}\: \\ $$$${difference}\:\mathrm{800}. \\ $$$$\Rightarrow{V}_{{n}} ={V}_{\mathrm{0}} +\mathrm{800}{n} \\ $$$${U}_{{n}+\mathrm{1}} −{U}_{{n}} ={V}_{\mathrm{0}} +\mathrm{800}{n} \\ $$$${U}_{{n}} −{U}_{{n}−\mathrm{1}} ={V}_{\mathrm{0}} +\mathrm{800}\left({n}−\mathrm{1}\right) \\ $$$$… \\ $$$${U}_{\mathrm{2}} −{U}_{\mathrm{1}} ={V}_{\mathrm{0}} +\mathrm{800}×\mathrm{1} \\ $$$${U}_{\mathrm{1}} −{U}_{\mathrm{0}} ={V}_{\mathrm{0}} +\mathrm{800}×\mathrm{0} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} −{U}_{\mathrm{0}} =\left({n}+\mathrm{1}\right){V}_{\mathrm{0}} +\mathrm{800}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} ={U}_{\mathrm{0}} +\left({n}+\mathrm{1}\right){V}_{\mathrm{0}} +\mathrm{800}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${or} \\ $$$${U}_{{n}} ={U}_{\mathrm{0}} +{nV}_{\mathrm{0}} +\mathrm{800}×\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{U}_{{n}} ={U}_{\mathrm{0}} +{n}\left({U}_{\mathrm{1}} −{U}_{\mathrm{0}} \right)+\mathrm{400}{n}\left({n}−\mathrm{1}\right) \\ $$$${with}\:{U}_{\mathrm{0}} =\mathrm{1},\:{U}_{\mathrm{1}} =\mathrm{2} \\ $$$$\Rightarrow{U}_{{n}} =\mathrm{1}+{n}+\mathrm{400}{n}\left({n}−\mathrm{1}\right) \\ $$
Commented by Ar Brandon last updated on 23/Nov/21
Thank you sir