let-U-n-a-sequence-U-0-a-and-U-n-nU-n-1-2-n-gt-0-calculate-U-n-interms-of-n-

Question Number 65134 by turbo msup by abdo last updated on 25/Jul/19

let U_n a sequence U_0 =a and U_n =nU_(n−1) −2 (n>0) calculate U_n interms of n.

$${let}\:{U}_{{n}} \:{a}\:{sequence}\:{U}_{\mathrm{0}} ={a}\:{and} \\ $$$${U}_{{n}} ={nU}_{{n}−\mathrm{1}} \:\:\:−\mathrm{2}\:\:\:\left({n}>\mathrm{0}\right) \\ $$$${calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n}. \\ $$

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!))

$$\:\:\:\:\:{Let}\:\:{U}_{{n}} =\:{n}!{V}_{{n}} \:\:\:{now}\:\:{let}\:{look}\:{for}\:{V}_{{n}} \\ $$$${the}\:{equality}\:\:{gives}\:\:{n}!{V}_{{n}\:} −{n}\left[\left({n}−\mathrm{1}\right){V}_{{n}−\mathrm{1}} \right]=−\mathrm{2}\: \\ $$$${then}\:\:\forall\:{k}\:>\mathrm{1}\:\:\:\:\:{V}_{{k}} −{V}_{{k}−\mathrm{1}} =\:\frac{−\mathrm{2}}{{k}!}\: \\ $$$${then}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({V}_{{k}} \:−{V}_{{k}−\mathrm{1}} \right)=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{−\mathrm{2}}{{k}!}\:\: \\ $$$$\:\:\:\:\:{finally}\:\:{V}_{{n}} \:=\:{V}_{\mathrm{0}} \:−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$$${V}_{\mathrm{0}} ={U}_{\mathrm{0}} ={a} \\ $$$${so}\:\:{U}_{{n}} =\:{a}.{n}!\:−{n}!\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{k}!} \\ $$

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

Commented by mathmax by abdo last updated on 25/Jul/19

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 26/Jul/19

let V_n =(U_n /(n!)) ⇒V_(n+1) −V_n =(U_(n+1) /((n+1)!)) −(U_n /(n!)) =(((n+1)U_n −2)/((n+1)!)) −(U_n /(n!)) =(U_n /(n!)) −(2/((n+1)!)) −(U_n /(n!)) =−(2/((n+1)!)) ⇒ Σ_(k=0) ^(n−1) (V_(k+1) −V_k ) =−2 Σ_(k=0) ^(n−1) (1/((k+1)!)) ⇒ V_1 −V_0 +V_2 −V_1 +....+V_n −V_(n−1) =−2 Σ_(k=1) ^n (1/(k!)) ⇒ V_n =V_0 −2Σ_(k=1) ^n (1/(k!)) =a −2 Σ_(k=1) ^n (1/(k!)) we have U_n =n!V_n ⇒ U_n =n!(a−2 Σ_(k=1) ^n (1/(k!))) remark we see that lim_(n→+∞) (U_n /(n!)) =a+2 −2e

$${let}\:{V}_{{n}} =\frac{{U}_{{n}} }{{n}!}\:\Rightarrow{V}_{{n}+\mathrm{1}} −{V}_{{n}} =\frac{{U}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!}\:=\frac{\left({n}+\mathrm{1}\right){U}_{{n}} −\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!} \\ $$$$=\frac{{U}_{{n}} }{{n}!}\:−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:−\frac{{U}_{{n}} }{{n}!}\:=−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({V}_{{k}+\mathrm{1}} −{V}_{{k}} \right)\:=−\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$${V}_{\mathrm{1}} −{V}_{\mathrm{0}} \:+{V}_{\mathrm{2}} −{V}_{\mathrm{1}} \:+….+{V}_{{n}} −{V}_{{n}−\mathrm{1}} =−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\Rightarrow \\ $$$${V}_{{n}} ={V}_{\mathrm{0}} \:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:={a}\:−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\:{we}\:{have}\:{U}_{{n}} ={n}!{V}_{{n}} \:\Rightarrow \\ $$$${U}_{{n}} ={n}!\left({a}−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\right) \\ $$$${remark}\:\:{we}\:{see}\:{that}\:{lim}_{{n}\rightarrow+\infty} \:\frac{{U}_{{n}} }{{n}!}\:={a}+\mathrm{2}\:−\mathrm{2}{e} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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