Question Number 114110 by bemath last updated on 17/Sep/20

(4/(1!)) + ((11)/(2!)) + ((22)/(3!)) + ((37)/(4!)) + ... = ?

Commented byDwaipayan Shikari last updated on 17/Sep/20

y △y △^2 y 4 7 11 4 11 22 4 15 37 φ(y)=4+7(n−1)+2(n−1)(n−2) =4+(n−1)(7+2n−4) =4+(n−1)(2n+3)=4+2n^2 +n−3=2n^2 +n+1 Σ^∞ ((2n^2 +n+1)/(n!))=4e+e+e−1=6e−1 Σ^∞ ((2n^2 )/(n!))=2((1^2 /(1!))+(2^2 /(2!))+(3^2 /(3!))+...)=2(1+(2/(1!))+(3/(2!))+....) =2(((1/(1!))+(2/(2!))+..)+(1+(1/(2!))+...))=2.2e=4e Σ^∞ (n/(n!))=(1/(1!))+(2/(2!))+.....=(1+(1/(1!))+(1/(2!))+....)=e Σ^∞ (1/(n!))=e−1

Commented bybemath last updated on 17/Sep/20

bravoo

Answered by bobhans last updated on 17/Sep/20

S = Σ_(n=1) ^∞ ((2n^2 +n+1)/(n!))=Σ_(n=1) ^∞ ((2n(n−1)+3n+1)/(n!)) = Σ_(n=1) ^∞ ((2n(n−1))/(n!))+Σ_(n=1) ^∞ ((3n)/(n!)) + Σ_(n=1) ^∞ (1/(n!)) S_1 =Σ_(n=1) ^∞ (1/(n!))=e−1. [ e^x =Σ_(n=0) ^∞ (x^n /(n!)) , letting x=1 ] S_2 =Σ_(n=1) ^∞ ((3n)/(n!)) = 3e . [ e^x =Σ_(n=1) ^∞ ((nx^(n−1) )/(n!)) ] S_3 = Σ_(n=1) ^∞ ((2n(n−1))/(n!)) = 2e. [ e^x =Σ_(n=1) ^∞ ((n(n−1)x^(n−2) )/(n!))] Hence S = S_1 +S_2 +S_3 = e−1+3e+2e= 6e−1