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Question Number 185542 by mathlove last updated on 23/Jan/23

Answered by Ar Brandon last updated on 23/Jan/23

u_n =3, 6, 10, 15, 21,... d_0 =3 Δu_n =3, 4, 5, 6,... d_1 =3 Δ^2 u_n =1, 1, 1, ... d_2 =1 u_n =d_0 +((d_1 (n−1))/(1!))+((d_2 (n−1)(n−2))/(2!)) =3+3(n−1)+(((n−1)(n−2))/2)=(n^2 /2)+((3n)/2)+1 S=1+(1/3)+(1/6)+(1/(10))+(1/(15))+∙∙∙=Σ_(n=0) ^∞ (2/(n^2 +3n+2)) =Σ_(n=0) ^∞ (2/((n+2)(n+1)))=Σ_(n=0) ^∞ ((2/(n+1))−(2/(n+2))) =2−lim_(n→∞) (2/(n+2))=2

$${u}_{{n}} =\mathrm{3},\:\mathrm{6},\:\mathrm{10},\:\mathrm{15},\:\mathrm{21},...\:{d}_{\mathrm{0}} =\mathrm{3} \\ $$$$\Delta{u}_{{n}} =\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},...\:{d}_{\mathrm{1}} =\mathrm{3} \\ $$$$\Delta^{\mathrm{2}} {u}_{{n}} =\mathrm{1},\:\mathrm{1},\:\mathrm{1},\:...\:{d}_{\mathrm{2}} =\mathrm{1} \\ $$$${u}_{{n}} ={d}_{\mathrm{0}} +\frac{{d}_{\mathrm{1}} \left({n}−\mathrm{1}\right)}{\mathrm{1}!}+\frac{{d}_{\mathrm{2}} \left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}!} \\ $$$$\:\:\:\:\:=\mathrm{3}+\mathrm{3}\left({n}−\mathrm{1}\right)+\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{n}}{\mathrm{2}}+\mathrm{1} \\ $$$${S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{15}}+\centerdot\centerdot\centerdot=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}} \\ $$$$\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}}{{n}+\mathrm{1}}−\frac{\mathrm{2}}{{n}+\mathrm{2}}\right) \\ $$$$\:\:\:=\mathrm{2}−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}}{{n}+\mathrm{2}}=\mathrm{2} \\ $$

Commented by mathlove last updated on 23/Jan/23

descraib to simple How is u_n =3,6,10,15,21.....d_0 =3 what′s d_0 and anuther one??

$${descraib}\:{to}\:{simple} \\ $$$${How}\:{is}\:\:\:\:{u}_{{n}} =\mathrm{3},\mathrm{6},\mathrm{10},\mathrm{15},\mathrm{21}.....{d}_{\mathrm{0}} =\mathrm{3}\:\:{what}'{s}\:{d}_{\mathrm{0}} \\ $$$${and}\:{anuther}\:{one}?? \\ $$

Commented by Ar Brandon last updated on 23/Jan/23

u_n is obtained using the terms at the denominator. d_0 is the first term.

$${u}_{{n}} \:\mathrm{is}\:\mathrm{obtained}\:\mathrm{using}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{at}\:\mathrm{the}\:\mathrm{denominator}. \\ $$$${d}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}. \\ $$

Answered by Frix last updated on 23/Jan/23

a_n =(2/(n(n+1))) Σ_(k=1) ^n a_k =2−(2/(n+1)) ⇒ answer is 2

$${a}_{{n}} =\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\mathrm{2}−\frac{\mathrm{2}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2} \\ $$

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