Question Number 104517 by Quvonchbek last updated on 22/Jul/20

Answered by 1549442205PVT last updated on 22/Jul/20

It is easy to see that the sequence of numbers:  11,19,29,41,...has general term is  n^2 +3n+1(n∈N^∗ ).Hence,  S=Σ_(n=1) ^(∞) ((n^2 +3n+1)/(n!))=Σ_(k=1) ^(∞) (n^2 /(n!))+Σ_(n=1) ^(∞) ((3n)/(n!))+Σ_(n=1) ^(∞) (1/(n!))(∗)  On the other hands,we have  e=Σ_(n=0) ^(∞) (1/(n!))=1+(1/(1!))+(1/(2!))+(1/(3!))+(1/(4!))+...Hence,  Σ_(n=1) ^∞ ((3n)/(n!))=(3/(1!))+((3.2)/(2!))+((3.3)/(3!))+((3.4)/(4!))+...  =3+(3/(1!))+(3/(2!))+(3/(3!))+...=3(1+(1/(1!))+(1/(2!))+(1/(3!))+...)  =3e(1).Also we have also  Σ_(n=1) ^∞ (1/(n!))=(1/(1!))+(1/(2!))+(1/(3!))+...=e−1(1)  Σ_(n=1) ^∞ (n^2 /(n!))=(1/(1!))+(4/(2!))+(9/(3!))+((16)/(4!))+...  =(1/(1!))+(2/(1!))+(3/(2!))+(4/(3!))+...  We have also that  e=1+(1/(1!))+(1/(2!))+(1/(3!))+(1/(4!))+...  =1+((2−1)/(1!))+((3−2)/(2!))+((4−3)/(3!))+((5−4)/(4!))+...  =(1/(1!))+(2/(1!))+(3/(2!))+(4/(3!))+(5/(4!))+...−(1+(1/(1!))+(1/(2!))+(1/(3!))+...)  =1+(2/(1!))+(3/(2!))+(4/(3!))+(5/(4!))+...−e  ⇒2e=(1/(1!))+(2/(1!))+(3/(2!))+(4/(3!))+(5/(4!))+...(2)  From(∗), (1)and(2) we get  S=Σ((n^2 +3n+1)/(n!))=e−1+3e+2e=6e−1  =ae^b +c⇒a=6,b=1,c=−1.Therefore,  a+b+c=6

Commented byOlafThorendsen last updated on 22/Jul/20

The sum starts at 1, not at 0.  Finally the good result is 6e−1 sir.

Commented by1549442205PVT last updated on 22/Jul/20

Thank you Sir.I mistaked and  corrected.

Answered by OlafThorendsen last updated on 22/Jul/20

S = Σ_(k=1) ^∞ (((k+1)(k+2)−1)/(k!))  S = Σ_(k=1) ^∞ ((k^2 +3k+1)/(k!))  S = Σ_(k=1) ^∞ (k/((k−1)!))+3Σ_(k=1) ^∞ (1/((k−1)!))+Σ_(k=1) ^∞ (1/(k!))  S = Σ_(k=0) ^∞ ((k+1)/(k!))+3Σ_(k=0) ^∞ (1/(k!))+Σ_(k=1) ^∞ (1/(k!))  S = Σ_(k=1) ^∞ (1/((k−1)!))+4Σ_(k=0) ^∞ (1/(k!))+Σ_(k=1) ^∞ (1/(k!))  S = Σ_(k=0) ^∞ (1/(k!))+4Σ_(k=0) ^∞ (1/(k!))+(Σ_(k=0) ^∞ (1/(k!))−1)  S = 6Σ_(k=0) ^∞ (1/(k!))−1  S = 6e−1

Commented byDwaipayan Shikari last updated on 22/Jul/20

😃

Commented byDwaipayan Shikari last updated on 22/Jul/20

Great solution

Commented bymr W last updated on 22/Jul/20

i agree with MJS sir. in fact the  sum of LHS is not unique, since the  a_n  term is not defined!

Commented by1549442205PVT last updated on 22/Jul/20

we can find out the rule to define   general term follows as:  a_1 =5,a_2 =11,a_3 =19,a_4 =29,a_5 =41.So,   determinant (((a_2 −a_1 ),(a_3 −a_2 ),(a_4 −a_3 ),(a_5 −a_4 )),(6,8,(10),(12)))  Since 12−10=10−8=8−6 ,it follows  that the sequence 6,8,10,12 ...form  an arithmetic progression which has  the difference equal to 2.Hence its  general term is 2n+4.So,by above result  a_(n+1) −a_n =2n+4.It follows that  Σ_(k=1) ^n (a_(k+1) −a_k )=Σ_(k=1) ^(n) (2k+4)  ⇒a_(n+1) −a_1 =2Σ_(k=1) ^(n) k+4n=n(n+1)+4n  ⇒a_(n+1) =n^2 +5n+5⇒a_n =n^2 +3n+1