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Question Number 59377 by jimful last updated on 09/May/19
Find Σ_(n=1) ^∞ ((1+2n+3n^2 +4n^3 )/3^n )
$$\mathrm{Find}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}+\mathrm{2n}+\mathrm{3n}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{n}} } \\ $$
Answered by tanmay last updated on 09/May/19
S=Σ_(n=1) ^∞ (1/3^n )+2Σ^∞ (n/3^n )+3Σ_(n=1) ^∞ (n^2 /3^n )+4Σ_n ^∞ (n^3 /3^n ) S=S_1 +2S_2 +3S_3 +4S_4 S_1 =((1/3)/(1−(1/3)))=(1/2) [formula S_1 =(a/(1−r))]←look here S_2 =(1/3^1 )+(2/3^2 )+(3/3^3 )+(4/3^4 )+... S_2 ×(1/3)= (1/3^2 )+(2/3^3 )+(3/3^4 )+... S_2 −S_2 ×(1/3)=(1/3)+(1/3^2 )+(1/3^3 )+... ((2S_2 )/3)=((1/3)/(1−(1/3)))=(1/2) so 2S_2 =(3/2)←look here S_3 =(1^2 /3^1 )+(2^2 /3^2 )+(3^2 /3^3 )+(4^2 /3^4 )+... ((S_3 ×(1/3)=(1/3^2 )+(2^2 /3^3 )+(3^2 /3^4 )+...)/) S_3 (1−(1/3))=(1/3^1 )+(3/3^2 )+(5/3^3 )+(7/3^4 )+... S_3 ×(2/3)×(1/3)= (1/3^2 )+(3/3^3 )+(5/3^4 )+.. S_3 ×(2/3)×(1−(1/3))=(1/3)+(2/3^2 )+(2/3^3 )+(2/3^4 )+... S_3 ×(4/9)=(1/3)+((2/3^2 )/(1−(1/3))) ((4S_3 )/9)=(1/3)+(2/3^2 )×(3/2) ((4S_3 )/9)=(2/3) 3S_3 =((2×9)/4)=(9/2)←look here S_4 =(1^3 /3^1 )+(2^3 /3^2 )+(3^3 /3^3 )+(4^3 /3^4 )+(5^3 /3^5 )+(6^3 /3^6 )+... S_4 ×(1/3)= (1/3^2 )+(2^3 /3^3 )+(3^3 /3^4 )+(4^3 /3^5 )+(5^3 /3^6 )+. S_4 (1−(1/3))=((1^3 −0^3 )/3^1 )+((2^3 −1^3 )/3^2 )+((3^3 −2^3 )/3^3 )+((4^3 −3^3 )/3^4 )+((5^3 −4^3 )/3^5 )+.. T_r =((r^3 −(r−1)^3 )/3^r ) =((r^3 −(r^3 −3r^2 +3r−1))/3^r ) S_4 ×(2/3)=Σ((3r^2 −3r+1)/3^r ) (2/3)×S_4 =Σ_(r=1) ^∞ [3((r^2 /3^r ))−3((r/3^r ))+((1/3^r ))] (2/3)×S_4 =3((3/2))−3((3/4))+(1/2) (2/3)×S_4 =5−(9/4) 4S_4 =((11×3)/2) so S=S_1 +2S_2 +3S_3 +4S_4 S=(1/2)+(3/2)+(9/2)+((33)/2)=((46)/2)=23 pls check mistake if any
$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{n}} }+\mathrm{2}\overset{\infty} {\sum}\frac{{n}}{\mathrm{3}^{{n}} }+\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }+\mathrm{4}\underset{{n}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} } \\ $$$${S}={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} +\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{4}} \\ $$$${S}_{\mathrm{1}} =\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[{formula}\:{S}_{\mathrm{1}} =\frac{{a}}{\mathrm{1}−{r}}\right]\leftarrow{look}\:{here} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{3}}=\:\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{2}} −{S}_{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+… \\ $$$$\frac{\mathrm{2}{S}_{\mathrm{2}} }{\mathrm{3}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:\mathrm{2}{S}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}\leftarrow{look}\:{here} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{4}} }+… \\ $$$$\frac{{S}_{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{4}} }+…}{} \\ $$$${S}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{7}}{\mathrm{3}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{3}} ×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{3}}=\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{4}} }+.. \\ $$$${S}_{\mathrm{3}} ×\frac{\mathrm{2}}{\mathrm{3}}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{3}} ×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{4}{S}_{\mathrm{3}} }{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{\mathrm{4}{S}_{\mathrm{3}} }{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{3}{S}_{\mathrm{3}} =\frac{\mathrm{2}×\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{2}}\leftarrow{look}\:{here} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{6}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{6}} }+… \\ $$$${S}_{\mathrm{4}} ×\frac{\mathrm{1}}{\mathrm{3}}=\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{5}} }+\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{6}} }+. \\ $$$${S}_{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}^{\mathrm{3}} −\mathrm{0}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{5}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{5}} }+.. \\ $$$${T}_{{r}} =\frac{{r}^{\mathrm{3}} −\left({r}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}^{{r}} }\:=\frac{{r}^{\mathrm{3}} −\left({r}^{\mathrm{3}} −\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}{r}−\mathrm{1}\right)}{\mathrm{3}^{{r}} } \\ $$$${S}_{\mathrm{4}} ×\frac{\mathrm{2}}{\mathrm{3}}=\Sigma\frac{\mathrm{3}{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{1}}{\mathrm{3}^{{r}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×{S}_{\mathrm{4}} =\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\mathrm{3}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{3}^{{r}} }\right)−\mathrm{3}\left(\frac{{r}}{\mathrm{3}^{{r}} }\right)+\left(\frac{\mathrm{1}}{\mathrm{3}^{{r}} }\right)\right] \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×{S}_{\mathrm{4}} =\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×{S}_{\mathrm{4}} =\mathrm{5}−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{4}{S}_{\mathrm{4}} =\frac{\mathrm{11}×\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{S}}={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} +\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{4}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{33}}{\mathrm{2}}=\frac{\mathrm{46}}{\mathrm{2}}=\mathrm{23} \\ $$$${pls}\:{check}\:{mistake}\:{if}\:{any} \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay last updated on 09/May/19
trying to solve another way... S_ =S_1 +2S_2 +3S_3 +4S_4 S_1 =Σ_(n=1) ^∞ (1/3^n ) S_2 =Σ_(n=1) ^∞ (n/3^n ) S_3 =Σ_(n=1) ^∞ (n^2 /3^n ) S_4 =Σ_(n=1) ^∞ (n^3 /3^n ) S_4 =(1^3 /3^1 )+(2^3 /3^2 )+(3^3 /3^3 )+(4^3 /3^4 )+... S_4 ×(1/3)=(1^3 /3^2 )+(2^3 /3^3 )+(3^3 /3^4 )+(4^3 /3^5 )+... S_4 (1−(1/3))=(1^3 /3^1 )+((2^3 −1^3 )/3^2 )+((3^3 −2^3 )/3^3 )+((4^3 −3^3 )/3^4 )+..+((n^3 −(n−1)^3 )/3^n )+.. S_4 ×(2/3)=Σ((3n^2 −3n+1)/3^n )=3Σ_(n=1) ^∞ (n^2 /3^n )−3Σ_(n=1) ^∞ (n/3^n )+Σ_(n=1) ^∞ (1/3^n ) (2/3)×S_4 =3S_3 −3S_2 +S_1 S_4 =((9S_3 )/2)−((9S_2 )/2)+((3S_1 )/2) S=S_1 +2S_2 +3S_3 +4S_4 S=S_1 +2S_2 +3S_3 +4×(((9S_3 )/2)−((9S_2 )/2)+((3S_1 )/2)) S=7S_1 −16S_2 +21S_3 S=7((1/2))−16((3/4))+21((3/2)) S=((70)/2)−((12)/1)=((46)/2)=23
$${trying}\:{to}\:{solve}\:{another}\:{way}… \\ $$$${S}_{} ={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} +\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{4}} \\ $$$${S}_{\mathrm{1}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{n}} } \\ $$$${S}_{\mathrm{2}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{3}^{{n}} } \\ $$$${S}_{\mathrm{3}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} } \\ $$$${S}_{\mathrm{4}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} } \\ $$$$\boldsymbol{{S}}_{\mathrm{4}} =\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+… \\ $$$${S}_{\mathrm{4}} ×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{5}} }+… \\ $$$${S}_{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+..+\frac{{n}^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}^{{n}} }+.. \\ $$$${S}_{\mathrm{4}} ×\frac{\mathrm{2}}{\mathrm{3}}=\Sigma\frac{\mathrm{3}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}^{{n}} }=\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}^{{n}} }−\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{3}^{{n}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{n}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×{S}_{\mathrm{4}} =\mathrm{3}{S}_{\mathrm{3}} −\mathrm{3}{S}_{\mathrm{2}} +{S}_{\mathrm{1}} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{9}{S}_{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{9}{S}_{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{S}_{\mathrm{1}} }{\mathrm{2}} \\ $$$${S}={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} +\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{4}} \\ $$$${S}={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} +\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}×\left(\frac{\mathrm{9}{S}_{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{9}{S}_{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{S}_{\mathrm{1}} }{\mathrm{2}}\right) \\ $$$${S}=\mathrm{7}{S}_{\mathrm{1}} −\mathrm{16}{S}_{\mathrm{2}} +\mathrm{21}{S}_{\mathrm{3}} \\ $$$${S}=\mathrm{7}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{16}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{21}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${S}=\frac{\mathrm{70}}{\mathrm{2}}−\frac{\mathrm{12}}{\mathrm{1}}=\frac{\mathrm{46}}{\mathrm{2}}=\mathrm{23} \\ $$
Commented by jimful last updated on 09/May/19
your answer is correct
$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by tanmay last updated on 10/May/19
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Answered by maxmathsup by imad last updated on 10/May/19
S =Σ_(n=1) ^∞ ((1/3))^n +2Σ_(n=1) ^∞ n((1/3))^n +3 Σ_(n=1) ^∞ n^2 ((1/3))^n +4Σ_(n=1) ^∞ n^3 ((1/3))^n =S_1 +2S_2 +3S_3 +4S_4 let S(x) =Σ_(n=0) ^∞ x^n with ∣x∣<1 ⇒S(x) =(1/(1−x)) ⇒S_1 =S((1/3))=(1/(1−(1/3))) =(3/2) S^′ (x) =Σ_(n=1) ^∞ nx^(n−1) ⇒xS^′ (x)=Σ_(n=1) ^∞ nx^n ⇒S_2 =(1/3) S^′ ((1/3)) but S^((1)) (x) =(1/((1−x)^2 )) ⇒S^((1)) ((1/3)) =(1/(((2/3))^2 )) =(9/4) ⇒S_2 =(1/3) (9/4) =(3/4) we have (x S^((1)) (x))^′ =Σ_(n=1) ^∞ n^2 x^(n−1) ⇒x (S^((1)) (x)+xS^((2)) (x))=Σ_(n=1) ^∞ n^2 x^n ⇒ S_3 =(1/3)( S^((1)) ((1/3))+(1/3) S^((2)) ((1/3)))=(1/3) (9/4) +(1/9) S^((2)) ((1/3)) but S^((2)) (x) = ((−2(−1)(1−x))/((1−x)^4 )) =(2/((1−x)^3 )) ⇒S^((2)) ((1/3)) =(2/(((2/3))^3 )) =2 ((27)/8) =((27)/4) ⇒ S_3 =(3/4) +(1/9) ((27)/4) =(3/4) +(3/4) =((2.3)/4) =(3/2) also we have xS^((1)) (x)+x^2 S^((2)) (x) =Σ_(n=1) ^∞ n^2 x^n ⇒S^((1)) (x)+xS^((2)) (x) +2x S^((2)) (x) +x^2 S^((3)) (x) =Σ_(n=1) ^∞ n^3 x^(n−1) ⇒x( S^((1)) (x) +3x S^((2)) (x) +x^2 S^((3)) (x))=Σ_(n=1) ^∞ n^3 x^n ⇒ S_4 =(1/3)( S^((1)) ((1/3))+S^((2)) ((1/3)) +(1/9) S^((3)) ((1/3))) but S^((1)) ((1/3)) =(9/4) , S^((2)) ((1/3)) =((27)/4) , S^((3)) (x) =((−2.3(−1)(1−x)^2 )/((1−x)^6 )) =(6/((1−x)^4 )) ⇒ S^((3)) ((1/3)) =(6/(((2/3))^4 )) =(6/((16)/(81))) =((6.81)/(16)) =((2.3.81)/(8.2)) =((243)/8) ⇒ S_4 =(1/3)( (9/4) +((27)/4) +(1/9) ((243)/8)) rest to achaive the calculus....
$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:+\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:+\mathrm{3}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} ={S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} \:+\mathrm{3}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{4}} \\ $$$${let}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{S}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{S}_{\mathrm{1}} ={S}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${S}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:\Rightarrow{xS}^{'} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:\Rightarrow{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:{S}^{'} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:{but} \\ $$$${S}^{\left(\mathrm{1}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow{S}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{1}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{9}}{\mathrm{4}}\:\Rightarrow{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{9}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${we}\:{have}\:\left({x}\:{S}^{\left(\mathrm{1}\right)} \left({x}\right)\right)^{'} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \:\Rightarrow{x}\:\left({S}^{\left(\mathrm{1}\right)} \left({x}\right)+{xS}^{\left(\mathrm{2}\right)} \left({x}\right)\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}} \:\Rightarrow \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\:{S}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{3}}\:{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{9}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{9}}\:{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:{but} \\ $$$${S}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\:\frac{−\mathrm{2}\left(−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }\:=\mathrm{2}\:\frac{\mathrm{27}}{\mathrm{8}}\:=\frac{\mathrm{27}}{\mathrm{4}}\:\Rightarrow \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{9}}\:\frac{\mathrm{27}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{2}.\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\:{also}\:{we}\:{have}\: \\ $$$${xS}^{\left(\mathrm{1}\right)} \left({x}\right)+{x}^{\mathrm{2}} {S}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}} \:\Rightarrow{S}^{\left(\mathrm{1}\right)} \left({x}\right)+{xS}^{\left(\mathrm{2}\right)} \left({x}\right)\:+\mathrm{2}{x}\:{S}^{\left(\mathrm{2}\right)} \left({x}\right)\:+{x}^{\mathrm{2}} \:{S}^{\left(\mathrm{3}\right)} \left({x}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} \:{x}^{{n}−\mathrm{1}} \:\Rightarrow{x}\left(\:{S}^{\left(\mathrm{1}\right)} \left({x}\right)\:+\mathrm{3}{x}\:{S}^{\left(\mathrm{2}\right)} \left({x}\right)\:+{x}^{\mathrm{2}} \:{S}^{\left(\mathrm{3}\right)} \left({x}\right)\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} \:{x}^{{n}} \:\Rightarrow \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\:{S}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{9}}\:{S}^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)\:\:{but} \\ $$$${S}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{9}}{\mathrm{4}}\:\:,\:\:{S}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{27}}{\mathrm{4}}\:,\:\:\:{S}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\frac{−\mathrm{2}.\mathrm{3}\left(−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} }\:=\frac{\mathrm{6}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$${S}^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{6}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{4}} }\:=\frac{\mathrm{6}}{\frac{\mathrm{16}}{\mathrm{81}}}\:=\frac{\mathrm{6}.\mathrm{81}}{\mathrm{16}}\:=\frac{\mathrm{2}.\mathrm{3}.\mathrm{81}}{\mathrm{8}.\mathrm{2}}\:=\frac{\mathrm{243}}{\mathrm{8}}\:\Rightarrow \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\:\frac{\mathrm{9}}{\mathrm{4}}\:+\frac{\mathrm{27}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{9}}\:\frac{\mathrm{243}}{\mathrm{8}}\right)\:{rest}\:{to}\:{achaive}\:{the}\:{calculus}…. \\ $$
Commented by jimful last updated on 11/May/19
brilliant solution
$$\mathrm{brilliant}\:\mathrm{solution} \\ $$$$ \\ $$

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