Question Number 138026 by mnjuly1970 last updated on 09/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:………{nice}\:\:…\:…\:…\:{calculus}……….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Theta=\:\underset{\overset{{n}=\mathrm{1}} {\:}} {\overset{\infty} {\sum}}\left(\frac{{n}^{\mathrm{2}} }{{n}!.\mathrm{4}^{{n}} }\right)\:=?? \\ $$
Answered by Dwaipayan Shikari last updated on 09/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}={e}^{{x}} −\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}} }{{n}!}={xe}^{{x}} \Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} {x}^{{n}} }{{n}!}={x}^{\mathrm{2}} {e}^{{x}} +{xe}^{{x}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!\mathrm{4}^{{n}} }=\frac{\mathrm{5}}{\mathrm{16}}{e}^{\mathrm{1}/\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 09/Apr/21
$$\:\:\:{thanks}\:{alot}\:…. \\ $$
Answered by mathmax by abdo last updated on 09/Apr/21
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\:\:=\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{nx}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}!}\: \\ $$$$\Rightarrow\mathrm{xf}^{'} \left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{nx}^{\mathrm{n}} }{\mathrm{n}!}\:\Rightarrow\left(\mathrm{xf}^{'} \left(\mathrm{x}\right)\right)^{'} \:=\Sigma\frac{\mathrm{n}^{\mathrm{2}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}!}\:\Rightarrow \\ $$$$\mathrm{x}\left(\mathrm{f}^{'} \left(\mathrm{x}\right)+\mathrm{xf}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\right)=\Sigma\:\frac{\mathrm{n}^{\mathrm{2}} \mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\:\Rightarrow\mathrm{x}\left(\mathrm{e}^{\mathrm{x}} \:+\mathrm{xe}^{\mathrm{x}} \right)\:=\Sigma\left(…\right)\:\Rightarrow \\ $$$$\Sigma\:\:\frac{\mathrm{n}^{\mathrm{2}} \mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}\right)\mathrm{e}^{\mathrm{x}} \:\:\:\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{n}!\mathrm{4}^{\mathrm{n}} }\:=\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{16}}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by mnjuly1970 last updated on 09/Apr/21
$${grateful}\:{sir} \\ $$
Answered by Ñï= last updated on 09/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!\mathrm{4}^{{n}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}+\mathrm{1}}{{n}!\mathrm{4}^{{n}} }=\frac{\mathrm{1}}{\mathrm{4}}\left({xD}+\mathrm{1}\right){e}^{{x}} \mid_{{x}=\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{1}\right){e}^{{x}} \mid_{{x}=\frac{\mathrm{1}}{\mathrm{4}}} =\frac{\mathrm{5}}{\mathrm{16}}{e}^{\mathrm{1}/\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 09/Apr/21
$$\:{tvhanks}\:{alot}.. \\ $$
Answered by mnjuly1970 last updated on 09/Apr/21
