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Question Number 78941 by M±th+et£s last updated on 21/Jan/20

Q.find the sum  S=(2^3 /(2!))+(3^3 /(3!))+(4^3 /(4!))+....  then find Σ_(n=1) ^∞ (n^4 /(n!))

Q.findthesumS=232!+333!+434!+....thenfindn=1n4n!

Answered by mind is power last updated on 21/Jan/20

=Σ_(n≥1) (n^3 /((n−1)!))=Σ_(k≥0) (((k+1)^3 )/(k!))=Σ_(k≥0) (((k^3 +3k^2 +3k+1))/(k!))  =Σ_(k≥1) (k^2 /((k−1)!))+3Σ_(k≥1) (k/((k−1)!))+3Σ_(k≥1) (1/((k−1)!))+Σ_(k≥1) (1/(k!))  =Σ_(k≥0) (((k+1)^2 )/(k!))+3Σ_(k≥0) ((k+1)/(k!))+3Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))−1  =Σ_(k≥0) ((k^2 +2k+1)/(k!))+3Σ_(k≥0) (k/(k!))+Σ_(k≥0) (7/(k!))−1  =Σ_(k≥1) (k/((k−1)!))+2Σ_(k≥1) (1/((k−1)!))+Σ_(k≥0) (1/(k!))+3Σ_(k≥1) (1/((k−1)!))+7Σ_(k≥0) (1/(k!))−1  =Σ_(k≥0) ((k+1)/(k!))+2Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))+10Σ_(k≥0) (1/(k!))−1  =Σ_(k≥1) (1/((k−1)!))+Σ_(k≥0) (1/(k!))+2Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))+10Σ_(k≥0) (1/(k!))−1  =15Σ_(k≥0) (1/(k!))−1=15e−1

=n1n3(n1)!=k0(k+1)3k!=k0(k3+3k2+3k+1)k!=k1k2(k1)!+3k1k(k1)!+3k11(k1)!+k11k!=k0(k+1)2k!+3k0k+1k!+3k01k!+k01k!1=k0k2+2k+1k!+3k0kk!+k07k!1=k1k(k1)!+2k11(k1)!+k01k!+3k11(k1)!+7k01k!1=k0k+1k!+2k01k!+k01k!+10k01k!1=k11(k1)!+k01k!+2k01k!+k01k!+10k01k!1=15k01k!1=15e1

Answered by Smail last updated on 22/Jan/20

Σ_(n=1) ^∞ (n^4 /(n!))=Σ_(n=1) ^∞ (n^3 /((n−1)(n−2)(n−3)(n−4)!))  (n^3 /((n−1)!))=(1/((n−1)!))+(7/((n−2)!))+(6/((n−3)!))+(1/((n−4)!))  S=Σ_(n=4) ^∞ ((1/((n−1)!))+(7/((n−2)!))+(6/((n−3)!))+(1/((n−4)!)))+(1+((16)/2)+((27)/2))  S=((45)/2)+Σ_(n=4) ^∞ (1/((n−1)!))+Σ_(n=4) ^∞ (7/((n−2)!))+Σ_(n=4) ^∞ (6/((n−3)!))+Σ_(n=4) ^∞ (1/((n−4)!))  S=((45)/2)+Σ_(n=3) ^∞ (1/(n!))+Σ_(n=2) ^∞ (7/(n!))+Σ_(n=1) ^∞ (6/(n!))+Σ_(n=0) ^∞ (1/(n!))  =((45)/2)+(e−1−1−(1/2))+7(e−1−1)+6(e−1)+e  =((45)/2)+15e−(5/2)−14−6  Σ_(n=1) ^∞ (n^4 /(n!))=15e

n=1n4n!=n=1n3(n1)(n2)(n3)(n4)!n3(n1)!=1(n1)!+7(n2)!+6(n3)!+1(n4)!S=n=4(1(n1)!+7(n2)!+6(n3)!+1(n4)!)+(1+162+272)S=452+n=41(n1)!+n=47(n2)!+n=46(n3)!+n=41(n4)!S=452+n=31n!+n=27n!+n=16n!+n=01n!=452+(e1112)+7(e11)+6(e1)+e=452+15e52146n=1n4n!=15e

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