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Question Number 65134 by turbo msup by abdo last updated on 25/Jul/19
$${let}\:{U}_{{n}} \:{a}\:{sequence}\:{U}_{\mathrm{0}} ={a}\:{and} \\ $$ $${U}_{{n}} ={nU}_{{n}−\mathrm{1}} \:\:\:−\mathrm{2}\:\:\:\left({n}>\mathrm{0}\right) \\ $$ $${calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n}. \\ $$
Commented by~ À ® @ 237 ~ last updated on 25/Jul/19
![Let U_n = n!V_n now let look for V_n the equality gives n!V_(n ) −n[(n−1)V_(n−1) ]=−2 then ∀ k >1 V_k −V_(k−1) = ((−2)/(k!)) then Σ_(k=1) ^n (V_k −V_(k−1) )= Σ_(k=1) ^n ((−2)/(k!)) finally V_n = V_0 −Σ_(k=1) ^n (2/(k!)) V_0 =U_0 =a so U_n = a.n! −n!Σ_(k=1) ^n (2/(k!))](Q65157.png)
$$\:\:\:\:\:{Let}\:\:{U}_{{n}} =\:{n}!{V}_{{n}} \:\:\:{now}\:\:{let}\:{look}\:{for}\:{V}_{{n}} \\ $$ $${the}\:{equality}\:\:{gives}\:\:{n}!{V}_{{n}\:} −{n}\left[\left({n}−\mathrm{1}\right){V}_{{n}−\mathrm{1}} \right]=−\mathrm{2}\: \\ $$ $${then}\:\:\forall\:{k}\:>\mathrm{1}\:\:\:\:\:{V}_{{k}} −{V}_{{k}−\mathrm{1}} =\:\frac{−\mathrm{2}}{{k}!}\: \\ $$ $${then}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({V}_{{k}} \:−{V}_{{k}−\mathrm{1}} \right)=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{−\mathrm{2}}{{k}!}\:\: \\ $$ $$\:\:\:\:\:{finally}\:\:{V}_{{n}} \:=\:{V}_{\mathrm{0}} \:−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$ $${V}_{\mathrm{0}} ={U}_{\mathrm{0}} ={a} \\ $$ $${so}\:\:{U}_{{n}} =\:{a}.{n}!\:−{n}!\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$
Commented by~ À ® @ 237 ~ last updated on 25/Jul/19
![Let U_n = n!V_n now let look for V_n the equality gives n!V_(n ) −n[(n−1)V_(n−1) ]=−2 then ∀ k >1 V_k −V_(k−1) = ((−2)/(k!)) then Σ_(k=1) ^n (V_k −V_(k−1) )= Σ_(k=1) ^n ((−2)/(k!)) finally V_n = V_0 −Σ_(k=1) ^n (2/(k!)) V_0 =U_0 =a so U_n = a.n! −n!Σ_(k=1) ^n (2/(k!))](Q65158.png)
$$\:\:\:\:\:{Let}\:\:{U}_{{n}} =\:{n}!{V}_{{n}} \:\:\:{now}\:\:{let}\:{look}\:{for}\:{V}_{{n}} \\ $$ $${the}\:{equality}\:\:{gives}\:\:{n}!{V}_{{n}\:} −{n}\left[\left({n}−\mathrm{1}\right){V}_{{n}−\mathrm{1}} \right]=−\mathrm{2}\: \\ $$ $${then}\:\:\forall\:{k}\:>\mathrm{1}\:\:\:\:\:{V}_{{k}} −{V}_{{k}−\mathrm{1}} =\:\frac{−\mathrm{2}}{{k}!}\: \\ $$ $${then}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({V}_{{k}} \:−{V}_{{k}−\mathrm{1}} \right)=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{−\mathrm{2}}{{k}!}\:\: \\ $$ $$\:\:\:\:\:{finally}\:\:{V}_{{n}} \:=\:{V}_{\mathrm{0}} \:−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$ $${V}_{\mathrm{0}} ={U}_{\mathrm{0}} ={a} \\ $$ $${so}\:\:{U}_{{n}} =\:{a}.{n}!\:−{n}!\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$
Commented bymathmax by abdo last updated on 25/Jul/19
$${thank}\:{you}\:{sir} \\ $$
Commented bymathmax by abdo last updated on 26/Jul/19
$${let}\:{V}_{{n}} =\frac{{U}_{{n}} }{{n}!}\:\Rightarrow{V}_{{n}+\mathrm{1}} −{V}_{{n}} =\frac{{U}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!}\:=\frac{\left({n}+\mathrm{1}\right){U}_{{n}} −\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!} \\ $$ $$=\frac{{U}_{{n}} }{{n}!}\:−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!}\:=−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$ $$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({V}_{{k}+\mathrm{1}} −{V}_{{k}} \right)\:=−\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:\Rightarrow \\ $$ $${V}_{\mathrm{1}} −{V}_{\mathrm{0}} \:+{V}_{\mathrm{2}} −{V}_{\mathrm{1}} \:+....+{V}_{{n}} −{V}_{{n}−\mathrm{1}} =−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\Rightarrow \\ $$ $${V}_{{n}} ={V}_{\mathrm{0}} \:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:={a}\:−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\:{we}\:{have}\:{U}_{{n}} ={n}!{V}_{{n}} \:\Rightarrow \\ $$ $${U}_{{n}} ={n}!\left({a}−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\right) \\ $$ $${remark}\:\:{we}\:{see}\:{that}\:{lim}_{{n}\rightarrow+\infty} \:\frac{{U}_{{n}} }{{n}!}\:={a}+\mathrm{2}\:−\mathrm{2}{e} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$