+0

# hard sum

0
97
3

Compute 1^2/2^1 + 2^2/2^2 + 3^2/2^3 + ... + n^2/2^n + ...

May 1, 2020

#1
0

sumfor(n, 1, 1000, n^2 / 2^n) = it converges to 6

May 1, 2020
#2
+30943
+3

"Compute 1^2/2^1 + 2^2/2^2 + 3^2/2^3 + ... + n^2/2^n + ..."

As follows:

May 1, 2020
#3
+25565
+2

Compute
$$\dfrac{1^2}{2^1} + \dfrac{2^2}{2^2} + \dfrac{3^2}{2^3} +~ \ldots~ + \dfrac{n^2}{2^n} +~ \ldots$$

$$\small{ \begin{array}{|rcllll|} \hline \text{Sum}&=& \mathbf{\dfrac{1^2}{2^1}} &\mathbf{+\dfrac{2^2}{2^2}} &\mathbf{+ \dfrac{3^2}{2^3}} &\mathbf{+ \dfrac{4^2}{2^4}} &\mathbf{+~ \ldots~ + \dfrac{n^2}{2^n} +~ \ldots } \\\\ &=& 1^2\left(\dfrac{1}{2}\right) &+2^2\left(\dfrac{1}{2}\right)^2 &+3^2\left(\dfrac{1}{2}\right)^3 &+4^2\left(\dfrac{1}{2}\right)^4 &+~ \ldots~ + n^2\left(\dfrac{1}{2}\right)^n \quad | \quad r = \dfrac{1}{2} \\\\ &=& 1^2r&+2^2r^2&+3^2r^3&+4^2r^4 &+~ \ldots~ + n^2r^n +~ \ldots~ \\\\ &=& 1r &+4r^2 &+9r^3 &+16r^4 &+~ \ldots~ \\\\ &=& 1r &+1r^2 &+1r^3 &+1r^4 &+~ \ldots~ \quad | \quad s_1=r(1+r+r^2+r^3+~ \ldots~ ) \\ & & &+3r^2 &+3r^3 &+3r^4 &+~ \ldots~ \quad | \quad s_2=3r^2(1+r+r^2+r^3+~ \ldots~ ) \\ & & & &+5r^3 &+5r^4 &+~ \ldots~ \quad | \quad s_3=5r^3(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &+7r^4 &+~ \ldots~ \quad | \quad s_4=7r^4(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &\ldots & +~ \ldots~ \quad | \quad \ldots \\ \hline \end{array} \\ \begin{array}{|rcllll|} \hline &=& \left(s_1+s_2+s_3+s_4+~ \ldots~\right) \\ &=& \left(r+3r^2+5r^3+7r^4+~ \ldots~\right) \left(1+r+r^2+r^3+~ \ldots~ \right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{1}{1-r}\left(r+3r^2+5r^3+7r^4+~ \ldots~\right) \\ \mathbf{\text{Sum}} &=& \mathbf{\dfrac{1}{1-r}S } \quad | \quad S= r+3r^2+5r^3+7r^4+~ \ldots \\ \hline \end{array} }$$

$$\small{ \begin{array}{|rcllll|} \hline \mathbf{S}&=& \mathbf{r} &\mathbf{+3r^2} &\mathbf{+5r^3} &\mathbf{+7r^4} &+~ \ldots \\ &=& r &+r^2 &+r^3 &+r^4 &+~ \ldots \quad | \quad S_1=r(1+r+r^2+r^3+~ \ldots~ ) \\ & & &+2r^2 &+2r^3 &+2r^4 &+~ \ldots \quad | \quad S_2=2r^2(1+r+r^2+r^3+~ \ldots~ ) \\ & & & &+2r^3 &+2r^4 &+~ \ldots \quad | \quad S_3=2r^3(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &+2r^4 &+~ \ldots \quad | \quad S_4=2r^4(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &\ldots& +~ \ldots~ \quad | \quad \ldots \\ \hline \end{array} \\ \begin{array}{|rcllll|} \hline &=& \left(S_1+S_2+S_3+S_4+~ \ldots~\right) \\ &=& \left(r+2r^2+2r^3+2r^4+~ \ldots~\right) \left(1+r+r^2+r^3+~ \ldots~ \right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{1}{1-r}\left(r+2r^2+2r^3+2r^4+~ \ldots~\right) \\ &=& \dfrac{r}{1-r}\left(1+2r+2r^2+2r^3+~ \ldots~\right) \\ &=& \dfrac{r}{1-r}\left(1+2r(1+r+r^2+r^3~ \ldots~)\right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{r}{1-r}\left(1+2r\left(\dfrac{1}{1-r}\right)\right) \\ &=& \dfrac{r}{1-r}\left(1+\dfrac{2r}{1-r}\right) \\ \mathbf{S}&=& \mathbf{\dfrac{r}{1-r}\left(\dfrac{1+r}{1-r}\right)} \\ \hline \end{array} }$$

$$\begin{array}{|rcll|} \hline \mathbf{\text{Sum}} &=& \mathbf{\left(\dfrac{1}{1-r}\right)S } \quad | \quad \mathbf{S=\dfrac{r}{1-r}\left(\dfrac{1+r}{1-r}\right)} \\\\ \text{Sum}&=&\left(\dfrac{1}{1-r}\right)\left(\dfrac{r}{1-r}\right)\left(\dfrac{1+r}{1-r}\right) \\\\ \text{Sum}&=&\dfrac{r(1+r)}{(1-r)^3} \quad | \quad r=\dfrac{1}{2} \\\\ \text{Sum}&=&\dfrac{\dfrac{1}{2}\left(1+\dfrac{1}{2}\right)}{\left(1-\dfrac{1}{2}\right)^3} \\\\ \text{Sum}&=&\dfrac{\dfrac{1}{2}\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{2}\right)^3} \\\\ \text{Sum}&=&\dfrac{\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{2}\right)^2} \\\\ \text{Sum}&=&\dfrac{\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{4}\right)} \\\\ \text{Sum}&=&\dfrac{12}{2} \\\\ \mathbf{\text{Sum}}&=& \mathbf{6} \\ \hline \end{array}$$

May 2, 2020