Menu Close

Find-the-value-of-n-1-n-2-1-n-2-x-n-n-




Question Number 61922 by Tawa1 last updated on 11/Jun/19
Find the value of:     Σ_(n = 1) ^∞   ((n^2  + 1)/(n + 2)). (x^n /(n!))
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}.\:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!} \\ $$
Commented by maxmathsup by imad last updated on 12/Jun/19
let p(x) =Σ_(n=1) ^∞  ((n^2 +1)/(n+2)) (x^(n+2) /(n!)) ⇒((p(x))/x^2 ) =Σ_(n=1) ^∞  ((n^2 +1)/(n+2)) (x^n /(n!))  let find p(x)  p^′ (x) =Σ_(n=1) ^∞ ((n^2 +1)/(n!)) x^(n+1)  =x{ Σ_(n=1) ^∞ (n^2 /(n!)) x^n  +Σ_(n=1) ^∞  (x^n /(n!))}  Σ_(n=1) ^∞  (x^n /(n!)) =e^x −1  Σ_(n=1) ^∞  (n^2 /(n!))x^n  =Σ_(n=1) ^∞  (n/((n−1)!))x^n  =Σ_(n=1) ^∞ ((n−1 +1)/((n−1)!)) x^n  =Σ_(n=1) ^∞ ((n−1)/((n−1)!))x^n  +Σ_(n=1) ^∞  (x^n /((n−1)!))  =Σ_(n=2) ^∞   (x^n /((n−2)!))  +Σ_(n=0) ^∞  (x^(n+1) /(n!)) =Σ_(n=0) ^∞   (x^(n+2) /(n!)) +Σ_(n=0) ^∞  (x^(n+1) /(n!)) =x^2 e^x   +x e^x   =(x^2  +x)e^x  ⇒p^′ (x) =x{ (x^2  +x)e^x  +e^x −1} =x{(x^2 +x+1)e^x −1}  =(x^3  +x^2  +x)e^x  −x ⇒p(x) =∫(x^3  +x^2  +x)e^x dx−(x^2 /2) +c  ∫xe^x  =xe^x −∫e^x  dx =xe^x −e^x  =(x−1)e^x   ∫ x^2 e^x dx =x^2 e^x  −∫2x e^x dx =x^2 e^x −2(x−1)e^x  =(x^2 −2x+2)e^x   ∫x^3 e^x dx  =x^3 e^x  −∫ 3x^2  e^x dx =x^3 e^x −3(x^2 −2x+2)e^x   =(x^3 −3x^2  +6x −6)e^x  ⇒  p(x) =(x^3 −3x^2  +6x−6 +x^2 −2x+2 +x−1)e^x  −(x^2 /2) +c  =(x^3 −2x^2 +5x −5)e^x  −(x^2 /2) +c  p(0) =0 =−5 +c ⇒c =5 ⇒p(x) =(x^3 −2x^2  +5x−5)e^x  −(x^2 /2) +5 ⇒  Σ_(n=1) ^∞  ((n^2 +1)/(n+2)) (x^n /(n!)) =((p(x))/x^2 ) =(x−2+(5/x) −(5/x^2 ))e^x −(1/2) +(5/x^2 )    (   x≠0) .
$${let}\:{p}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{2}}\:\frac{{x}^{{n}+\mathrm{2}} }{{n}!}\:\Rightarrow\frac{{p}\left({x}\right)}{{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{2}}\:\frac{{x}^{{n}} }{{n}!}\:\:{let}\:{find}\:{p}\left({x}\right) \\ $$$${p}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}!}\:{x}^{{n}+\mathrm{1}} \:={x}\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}^{\mathrm{2}} }{{n}!}\:{x}^{{n}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\right\} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\:={e}^{{x}} −\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} }{{n}!}{x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\left({n}−\mathrm{1}\right)!}{x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}−\mathrm{1}\:+\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:{x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{n}−\mathrm{1}}{\left({n}−\mathrm{1}\right)!}{x}^{{n}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}−\mathrm{1}\right)!} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}} }{\left({n}−\mathrm{2}\right)!}\:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}+\mathrm{2}} }{{n}!}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}!}\:={x}^{\mathrm{2}} {e}^{{x}} \:\:+{x}\:{e}^{{x}} \\ $$$$=\left({x}^{\mathrm{2}} \:+{x}\right){e}^{{x}} \:\Rightarrow{p}^{'} \left({x}\right)\:={x}\left\{\:\left({x}^{\mathrm{2}} \:+{x}\right){e}^{{x}} \:+{e}^{{x}} −\mathrm{1}\right\}\:={x}\left\{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){e}^{{x}} −\mathrm{1}\right\} \\ $$$$=\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \:+{x}\right){e}^{{x}} \:−{x}\:\Rightarrow{p}\left({x}\right)\:=\int\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \:+{x}\right){e}^{{x}} {dx}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c} \\ $$$$\int{xe}^{{x}} \:={xe}^{{x}} −\int{e}^{{x}} \:{dx}\:={xe}^{{x}} −{e}^{{x}} \:=\left({x}−\mathrm{1}\right){e}^{{x}} \\ $$$$\int\:{x}^{\mathrm{2}} {e}^{{x}} {dx}\:={x}^{\mathrm{2}} {e}^{{x}} \:−\int\mathrm{2}{x}\:{e}^{{x}} {dx}\:={x}^{\mathrm{2}} {e}^{{x}} −\mathrm{2}\left({x}−\mathrm{1}\right){e}^{{x}} \:=\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right){e}^{{x}} \\ $$$$\int{x}^{\mathrm{3}} {e}^{{x}} {dx}\:\:={x}^{\mathrm{3}} {e}^{{x}} \:−\int\:\mathrm{3}{x}^{\mathrm{2}} \:{e}^{{x}} {dx}\:={x}^{\mathrm{3}} {e}^{{x}} −\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right){e}^{{x}} \\ $$$$=\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{6}{x}\:−\mathrm{6}\right){e}^{{x}} \:\Rightarrow \\ $$$${p}\left({x}\right)\:=\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{6}{x}−\mathrm{6}\:+{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\:+{x}−\mathrm{1}\right){e}^{{x}} \:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c} \\ $$$$=\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}\:−\mathrm{5}\right){e}^{{x}} \:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c} \\ $$$${p}\left(\mathrm{0}\right)\:=\mathrm{0}\:=−\mathrm{5}\:+{c}\:\Rightarrow{c}\:=\mathrm{5}\:\Rightarrow{p}\left({x}\right)\:=\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{5}{x}−\mathrm{5}\right){e}^{{x}} \:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{5}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{2}}\:\frac{{x}^{{n}} }{{n}!}\:=\frac{{p}\left({x}\right)}{{x}^{\mathrm{2}} }\:=\left({x}−\mathrm{2}+\frac{\mathrm{5}}{{x}}\:−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\right){e}^{{x}} −\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\:\:\:\:\left(\:\:\:{x}\neq\mathrm{0}\right)\:. \\ $$$$ \\ $$
Commented by Tawa1 last updated on 12/Jun/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Mr X pcx last updated on 12/Jun/19
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by Smail last updated on 12/Jun/19
((n^2 +1)/(n+2))=(((n+2)(n−2)+5)/(n+2))=n−2+(5/(n+2))  Let f(x)=Σ_(n=1) ^∞ ((n^2 +1)/(n+2))×(x^n /(n!))  f(x)=Σ_(n=1) ^∞ (n−2+(5/(n+2)))(x^n /(n!))  =Σ_(n=1) ^∞ (n/(n!))x^n −2Σ_(n=1) ^∞ (x^n /(n!))+5Σ_(n=1) ^∞ (x^n /((n+2)n!))  =Σ_(n=0) ^∞ (x^(n+1) /(n!))−2(Σ_(n=0) ^∞ (x^n /(n!))−1)+(5/x^2 )Σ_(n=1) ^∞ (x^(n+2) /((n+2)n!))  =xe^x −2(e^x −1)+(5/x^2 )p(x)  p(x)=Σ_(n=1) ^∞ (x^(n+2) /((n+2)n!))  p′(x)=Σ_(n=1) ^∞ (x^(n+1) /(n!))=x(e^x −1)  p(x)=∫_0 ^x (te^t −t)dt=[e^t (t−1)−(t^2 /2)]_0 ^x   p(x)=e^x (x−1)−(x^2 /2)+1  f(x)=xe^x −2e^x +2+(5/x^2 )(e^x (x−1)−(x^2 /2)+1)  f(x)=Σ_(n=1) ^∞ ((n^2 +1)/((n+2)n!))x^n =e^x (x−2+((5(x−1))/x^2 ))+(5/x^2 )−(1/2)
$$\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{2}}=\frac{\left({n}+\mathrm{2}\right)\left({n}−\mathrm{2}\right)+\mathrm{5}}{{n}+\mathrm{2}}={n}−\mathrm{2}+\frac{\mathrm{5}}{{n}+\mathrm{2}} \\ $$$${Let}\:{f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}+\mathrm{2}}×\frac{{x}^{{n}} }{{n}!} \\ $$$${f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{2}+\frac{\mathrm{5}}{{n}+\mathrm{2}}\right)\frac{{x}^{{n}} }{{n}!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{n}!}{x}^{{n}} −\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}+\mathrm{5}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{\left({n}+\mathrm{2}\right){n}!} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}!}−\mathrm{2}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}−\mathrm{1}\right)+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{2}} }{\left({n}+\mathrm{2}\right){n}!} \\ $$$$={xe}^{{x}} −\mathrm{2}\left({e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }{p}\left({x}\right) \\ $$$${p}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{2}} }{\left({n}+\mathrm{2}\right){n}!} \\ $$$${p}'\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}!}={x}\left({e}^{{x}} −\mathrm{1}\right) \\ $$$${p}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \left({te}^{{t}} −{t}\right){dt}=\left[{e}^{{t}} \left({t}−\mathrm{1}\right)−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{x}} \\ $$$${p}\left({x}\right)={e}^{{x}} \left({x}−\mathrm{1}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{1} \\ $$$${f}\left({x}\right)={xe}^{{x}} −\mathrm{2}{e}^{{x}} +\mathrm{2}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\left({e}^{{x}} \left({x}−\mathrm{1}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}+\mathrm{2}\right){n}!}{x}^{{n}} ={e}^{{x}} \left({x}−\mathrm{2}+\frac{\mathrm{5}\left({x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} }\right)+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 12/Jun/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *