$$\sum_{k=1}^{\infty}\frac{k^2}{k!}=?$$ Serinin toplami kactir?

Question

$$\sum_{k=1}^{\infty}\frac{k^2}{k!}=?$$

Comments

Lutfen neler denediginizi de ekleyiniz, Okkes hocam. Tesekkurler.

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2e olabilir mi?

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Soru, ahım şahım çözemeyeceginiz bir soru olmadıgından, siteye katkı için sordugunuzdan gereklı acıklamayı ve cozumu ekledım.

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Evet @eynesi  cevap 2e. Eger cozumunuz farkli ise paylasabilirmisiniz?

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aslında bent sayılarıyla bır şey cıkıyor tam buna cuk dıye oturan bır ara atar bırı :)

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Ben yukarda bosluga konusmusum gibi olmus. Admin de onemsememis pek.

Answer 1

1. Aşama:

$$\displaystyle\sum_{k=0}^\infty\dfrac{(k+1)^2}{k!}=1+\sum_{k=1}^\infty\dfrac{k^2+2k+1  }{ k!}$$ $$=$$ $$1+\sum_{k=1}^\infty\dfrac{k  }{ (k-1)!}+2\sum_{k=1}^\infty\dfrac{ 1 }{ (k-1)!}+\color{green}{\sum_{k=1}^\infty\dfrac{  1}{ k!}}$$ $$=$$ $$\sum_{k=1}^\infty\dfrac{k +1-1 }{ (k-1)!}+2\sum_{k=0}^\infty\dfrac{ 1 }{ k!}+\color{green}{\sum_{k=1}^\infty\dfrac{  1}{ k!}}$$ $$1+\displaystyle\sum_{k=2}^\infty\dfrac1{(k-2)!}+\sum_{k=2}^\infty\dfrac{1}{(k-1)!}+\color{green}{3\sum_{k=0}^\infty\dfrac{  1}{ k!}}$$ $$=$$ $$\color{red}{\boxed{\boxed{\displaystyle\sum_{k=0}^\infty\dfrac{(k+1)^2}{k!}=5\sum_{k=0}^\infty\dfrac1{k!}}}}$$

$$-----------------------$$

2. Aşama: 

2,a:

$$2\displaystyle\sum_{k=0}^\infty\dfrac{k}{k!}=2\displaystyle\sum_{k=1}^\infty\dfrac{k}{k!}=2\displaystyle\sum_{k=1}^\infty\dfrac{1}{(k-1)!}=2\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}$$ $$Yani\quad \boxed{2\displaystyle\sum_{k=0}^\infty\dfrac{k}{k!}=2\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}}$$

$$-----------------------$$

2,b:

$$\displaystyle\sum_{k=0}^\infty\dfrac{(k+1)^2}{k!}=\color{darkblue}{\displaystyle\sum_{k=0}^\infty\dfrac{k^2}{k!}}+\displaystyle\sum_{k=0}^\infty\dfrac{2k}{k!}+\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}$$ $$\Longrightarrow$$ $$\boxed{\boxed{\displaystyle\sum_{k=0}^\infty\dfrac{(k+1)^2}{k!}=\color{darkblue}{\displaystyle\sum_{k=0}^\infty\dfrac{k^2}{k!}}+3\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}}}$$  

$$-----------------------$$

2,c:

$$\boxed{\boxed{e^x=\displaystyle\sum_{k=0}^\infty\dfrac{x^k}{k!}\quad\to\quad e=\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}}}$$

$$-----------------------$$
Olduğundan;

$$\displaystyle\sum_{k=0}^\infty\dfrac{(k+1)^2}{k!}=5\sum_{k=0}^\infty\dfrac1{k!}=\color{darkblue}{\displaystyle\sum_{k=0}^\infty\dfrac{k^2}{k!}}+3\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}$$
$$\to$$ $$\color{purple}{\boxed{\boxed{\boxed{\displaystyle\sum_{k=0}^\infty\dfrac{k^2}{k!}=2\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}=2e}}}}$$  

Answer 2

Daha öz bir metod:


$$\displaystyle\sum_{k=1}^\infty\dfrac{k^2}{k!}=\displaystyle\sum_{k=1}^\infty\dfrac{k(k-1)}{k!}+\displaystyle\sum_{k=1}^\infty\dfrac{k}{k!}$$


$$\displaystyle\sum_{k=1}^\infty\dfrac{k(k-1)}{k!}=\displaystyle\sum_{k=2}^\infty\dfrac{k(k-1)}{k!}=\displaystyle\sum_{k=2}^\infty\dfrac{1}{(k-2)!}=\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}\tag1$$

$$Ve$$


$$\displaystyle\sum_{k=1}^\infty\dfrac{k}{k!}=\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}$$


$$Ve$$

$$e=\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}$$  Olduğundan;

$$\displaystyle\sum_{k=1}^\infty\dfrac{k^2}{k!}=\displaystyle\sum_{k=1}^\infty\dfrac{k(k-1)}{k!}+\displaystyle\sum_{k=1}^\infty\dfrac{k}{k!}=2\displaystyle\sum_{k=0}^\infty\dfrac{1}{k!}=2e$$

Comments

quzel bir yol..

Answer 3

Ben soyle cozmustum:

$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$   her iki tarafin  $x$  gore iki defa turevini alirsak ve     $x=1$   dersek

$$e=\sum_{k=0}^{\infty}\frac{k(k-1)}{k!}=\sum_{k=0}^{\infty}\frac{k^2}{k!}-\sum_{k=0}^{\infty}\frac{k}{k!}$$  

$$e+\sum_{k=0}^{\infty}\frac{k}{k!}=\sum_{k=0}^{\infty}\frac{k^2}{k!}$$    ilk terimler 0 oldugundan

$$e+\sum_{k=1}^{\infty}\frac{k}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{k!}$$

$$e+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}=\sum_{k=1}^{\infty}\frac{k^2}{k!}$$

$$e+\sum_{k=0}^{\infty}\frac{1}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{k!}$$

$$e+e=\sum_{k=1}^{\infty}\frac{k^2}{k!}$$